#  Orthogonal inner product graphs of odd characteristic and their
automorphisms††thanks:  Supported by the National Natural Science Foundation
of China (No. 11371343), Guangxi science and technology development foundation
(No. 1599005-2-13), the Guangxi Science Foundation (No. 0832107, 0640070,
0991102) and the Scientific Research Foundation of Guangxi Educational
Committee.

Shouxiang ZHAOa,b, Hengbin ZHANGc, Jizhu NANa, Gaohua TANGd Corresponding
author. E-mail<EMAIL_ADDRESS>
(aSchool of Mathematical Sciences, Dalian University of Technology, Dalian
116024, P.R. China
bDepartment of Mathematics and Computer Science, Guilin Normal College, Guilin
541001, P.R. China
cCollege of Mathematical Science, Yangzhou University, Yangzhou 225002, P.R.
China
dSchool of Mathematical Sciences, Guangxi Teachers Education University,
Nanning, 530023, P.R. China)

Abstract: Let $\mathbb{F}_{q}$ be a finite field of odd characteristic and
$2\nu+\delta\geq 2$ an integer number with $\delta=0,1$ or $2$. The orthogonal
inner product graph $Oi\big{(}2\nu+\delta,q\big{)}$ over $\mathbb{F}_{q}$ is
defined and the automorphism groups of $Oi\big{(}2\nu+\delta,q\big{)}$ are
determined. We show that $Oi\big{(}2\nu+\delta,q\big{)}$ is a disconnected
graph if $2\nu+\delta=2$; otherwise it is not. Moreover, we have two necessary
and sufficient conditions for two vertices of $Oi\big{(}2\nu+\delta,q\big{)}$
and two edges of $Oi\big{(}2\nu+\delta,q\big{)}$ respectively are in the same
orbit under the action of the automorphism group of
$Oi\big{(}2\nu+\delta,q\big{)}.$

Keywords: Graph; Orthogonal group; Orthogonal subspace; Automorphism

MSC (2010): 05C25, 05C60, 15A63

##  1 Introduction

Let $\mathbb{F}_{q}$ be a finite field of odd characteristic and $n\geq 1$ an
integer number. Let
$\mathbb{F}_{q}^{(n)}=\\{(a_{1},\ldots,a_{n}):a_{i}\in\mathbb{F}_{q}\ {\rm
for}\ i=1,\ldots,n\\}$ be the $n$-dimensional vector space over
$\mathbb{F}_{q}$. For the $m$-dimensional vector subspace $P$ of
$\mathbb{F}_{q}^{(n)}$, any $m\times n$ matrix whose rows,
$\alpha_{1},\ldots,\alpha_{m}$, form a basis of the space $P$, is called a
_matrix representation_ of the space $P$, which is also denoted by $P$. To
simplify notation, sometimes we write $[\alpha_{1},\ldots,\alpha_{m}]$ instead
of the matrix $P$. Let
$e_{1}=(1,0,\ldots,0),e_{2}=(0,1,\ldots,0),\ldots,e_{n}=(0,0,\ldots,1)$. Then
$e_{i},i=1,2,\ldots,n$, form a basis of $\mathbb{F}_{q}^{(n)}$. The set of all
non-zero elements of $\mathbb{F}_{q}$, denoted by $\mathbb{F}_{q}^{*}$, form a
cyclic group under field multiplication, called the _multiplicative group_ of
$\mathbb{F}_{q}$. The set of all square elements of $\mathbb{F}_{q}^{*}$,
denoted by $\mathbb{F}_{q}^{*2}$, is a subgroup of $\mathbb{F}_{q}^{*}$. Let
$z$ be a fixed non-square element in $\mathbb{F}_{q}^{*}$ and

$S_{2\nu+\delta,\Delta}=\left(\begin{array}[]{ccc}0&I^{(\nu)}&\\\
I^{(\nu)}&0&\\\ &&\Delta\end{array}\right),$

be four $(2\nu+\delta)\times(2\nu+\delta)$ nonsingular symmetric matrices,
where

$\Delta=\left\\{\begin{array}[]{ll}\emptyset,&\textrm{if $\delta=0,$}\\\ (1)\
{\rm or}\ (z),&\textrm{if $\delta=1,$}\\\ {\rm diag}(1,-z),&\textrm{if
$\delta=2.$}\end{array}\right.$

To simplify notations, we omit $\emptyset$ when $\delta=0$ and write $S$ for
$S_{2\nu+\delta,\Delta}$. A $(2\nu+\delta)\times(2\nu+\delta)$ nonsingular
matrix $T$ over $\mathbb{F}_{q}$ is called _orthogonal_ with respect to $S$ if
$TS\,^{t}\\!T=S$. The set of all $(2\nu+\delta)\times(2\nu+\delta)$ orthogonal
matrices form a group under matrix multiplication, called the _orthogonal
group_ of degree $2\nu+\delta$ with respect to $S$ over $\mathbb{F}_{q}$ and
denoted by $O_{2\nu+\delta}\big{(}\mathbb{F}_{q}\big{)}$. The factor group
$O_{2\nu+\delta}\big{(}\mathbb{F}_{q}\big{)}\big{/}\big{\\{}I,-I\big{\\}}$ is
called the _projective orthogonal group_ of degree $2\nu+\delta$ over
$\mathbb{F}_{q}$ and denoted by
$PO_{2\nu+\delta}\big{(}\mathbb{F}_{q}\big{)}$. There is an action of
$O_{2\nu+\delta}(\mathbb{F}_{q})$ on $\mathbb{F}_{q}^{(2\nu+\delta)}$ defined
as follows:

$\displaystyle\mathbb{F}_{q}^{(2\nu+\delta)}\times
O_{2\nu+\delta}(\mathbb{F}_{q})$ $\displaystyle\longrightarrow$
$\displaystyle\mathbb{F}_{q}^{(2\nu+\delta)}$
$\displaystyle((x_{1},\ldots,x_{2\nu+\delta}),T)$
$\displaystyle\longrightarrow$ $\displaystyle(x_{1},\ldots,x_{2\nu+\delta})T.$

The vector space $\mathbb{F}_{q}^{(2\nu+\delta)}$ together with this group
action is said to be the $(2\nu+\delta)$-dimensional _orthogonal space_ over
$\mathbb{F}_{q}.$ The map:
$\mathbb{F}_{q}^{(2\nu+\delta)}\times\mathbb{F}_{q}^{(2\nu+\delta)}\to\mathbb{F}_{q}$
given by $(\alpha,\beta)\mapsto\alpha S\,{{}^{t}\\!\beta}$ is said to be an
_orthogonal inner product_ of $\mathbb{F}_{q}^{(2\nu+\delta)}$ with respect to
$S$. For any subspace $P$ of $\mathbb{F}_{q}^{(2\nu+\delta)}$, the set
$\\{\alpha\in\mathbb{F}_{q}^{(2\nu+\delta)}:\alpha S\,^{t}\\!\beta=0$ for all
$\beta\in P\\}$, denoted by $P^{\perp}$, is said to be the _dual subspace_ of
$P$ with respect to $S$.

In general, determining all the automorphisms of a graph is an important but
difficult problem both in graph theory and in algebraic theory. In 2006, Tang
and Wan [4] introduced the concept of the symplectic graph over a finite
field. They proved that the symplectic graph is strongly regular and
determined its automorphism group. In [3, 7, 8], the authors introduced the
concepts of the orthogonal graph and the unitary graph over a finite field and
determined their automorphism groups, respectively. In 2014, Wong et al. [5]
introduced the concept of the zero-divisor graph based on rank one upper
triangular matrices and also determined its automorphism group. Motivated by
previous studies, we introduce a new graph on non-trivial orthogonal spaces
over a finite field for studying the interplay between properties of
orthogonal subspaces and the structure of graphs.

The _orthogonal inner product graph_ with relative to $S$ over
$\mathbb{F}_{q}$, denoted by $Oi\big{(}2\nu+\delta,q\big{)}$, is the graph
defined on all non-trivial orthogonal subspaces of
$\mathbb{F}_{q}^{(2\nu+\delta)}$ with an edge between vertices $A$ and $B,$
denoted by $A$—$B,$ if and only if $AS\,^{t}\\!B=0$. The set of all vertices
and all edges of $Oi\big{(}2\nu+\delta,q\big{)}$ are denoted by
$V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ and
$E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)},$ respectively. Of course, if
$A$—$B$ is in $E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$, then $B$ is a
subspace of $A^{\perp}$ and $A$ is a subspace of $B^{\perp}$.

In section 2, we show that $Oi\big{(}2\nu+\delta,q\big{)}$ is connected with
diameter $4$ when $2\nu+\delta\geq 3$ and also show that
$Oi\big{(}2\nu_{1}+\delta_{1},q_{1}\big{)}\cong
Oi\big{(}2\nu_{2}+\delta_{2},q_{2}\big{)}$ if and only if $\nu_{1}=\nu_{2},$
$\delta_{1}=\delta_{2}$ and $q_{1}=q_{2}$. In section 3, we obtain two
necessary and sufficient conditions: (1) two vertices of
$V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ are in the same orbit under the
action of ${\rm Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ if and only
if they are the same type subspace; (2) two edges $g$ and $f$ of
$E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ are in the same orbit under the
action of ${\rm Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ if and only
if the following conditions hold: (i) one vertex of $g$ and one vertex of $f$
are the same type subspace, (ii) the other vertex of $g$ and the other vertex
of $f$ are the same type subspace, and (iii) the sum of the two vertices of
$g$ and the sum of the two vertices of $f$ are also the same type subspace.
Furthermore, we show that ${\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}=PO_{2\nu+\delta}\big{(}\mathbb{F}_{q}\big{)}\cdot
E_{2\nu+\delta}$, where $E_{2\nu+\delta}$ is a subgroup of ${\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$.

##  2 Some properties of orthogonal inner graphs

In [2], a graph $G$ is _connected_ if there is a path between any two vertices
of $G$; otherwise $G$ is disconnected. The _distance_ between vertices $a$ and
$b$ in $G$ is the number of edges in a shortest path between $a$ and $b$ and
denoted by $d(a,b)$. The _diameter_ of G is the greatest distance between any
two vertices of $G$ and denoted by ${\rm diam}(G)$. The _degree_ of $x\in G$
is defined to be the number of edges of the form $x$—$y$ in $G$ and denoted by
$\deg(x)$.

######  Theorem 2.1.

Let $\mathbb{F}_{q}$ be a finite field. Then $Oi\big{(}2\nu+\delta,q\big{)}$
is a connected graph if and only if $2\nu+\delta\geq 3$. Moreover, if
$Oi\big{(}2\nu+\delta,q\big{)}$ is a connected graph, then
diam$\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}=4$.

###### Proof.

In order to prove the theorem, we need only consider three cases: $\delta=0,1$
and $2$. Without loss of generality, we need only consider the case of
$\delta=0$. Other cases are similar to prove. Suppose that $\delta=0$.

Let $\delta=0$ and $\nu=1$. Clearly, orthogonal subspaces $[e_{1}]$ and
$[e_{2}]$ of $\mathbb{F}_{q}^{(2)}$ are different. Since
$[e_{1}]=[e_{1}]^{\bot}$, $[e_{2}]=[e_{2}]^{\bot}$ and
$[e_{1},e_{2}]^{\bot}=0$, we have d$([e_{1}],[e_{2}])=\infty$ and so
$Oi\big{(}2,q\big{)}$ is not connected.

Let $\delta=0$, $\nu\geq 2$ and $A,B\in
V\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$. If $A$ and $B$ are adjacent, then
$A$—$B$ is a path of length 1. So in the remainder of the proof we assume that
$A$ and $B$ are not adjacent. Then there exist $[\alpha],[\beta]\in
V\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$ such that $[\alpha]\subseteq A^{\perp}$
and $[\beta]\subseteq B^{\perp}$. If $[\alpha]$ and $[\beta]$ are adjacent,
then $A$— $[\alpha]$—$[\beta]$—$B$ is a path of length 3. If $[\alpha]$ and
$[\beta]$ are not adjacent, then $\dim([\alpha,\beta]^{\perp})\geq 2\nu-2>2$
and there exists $[\gamma]\subseteq[\alpha,\beta]^{\perp}$ such that
$A$—$[\alpha]$—$[\gamma]$—$[\beta]$—$B$ is a path of length 4. Therefore
$Oi\big{(}2\nu,q\big{)}$ is connected and
diam$\big{(}Oi\big{(}2\nu,q\big{)}\big{)}\leq 4$. Let
$A=[e_{1},\ldots,e_{2\nu-1}]$ and
$B=[e_{1},\ldots,e_{\nu-1},e_{\nu+1},\ldots,e_{2\nu}]$ with
$A^{\bot}=[e_{\nu}]$ and $B^{\bot}=[e_{2\nu}]$. Then
$A$—$[e_{\nu}]$—$[e_{1}]$—$[e_{2\nu}]$—$B$ is a shortest path of length 4
between $A$ and $B$ and so diam$\big{(}Oi\big{(}2\nu,q\big{)}\big{)}=4$.

The set of all vertices of dimension 1 in $Oi\big{(}2\nu+\delta,q\big{)}$ can
induce a subgraph of $Oi\big{(}2\nu+\delta,q\big{)}$, denoted by
$Oi\big{(}2\nu+\delta,1,q\big{)}$, with an edge between vertices $A$ and $B$
if and only if $AS\,^{t}\\!B=0.$ Of course, if $X\in
V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}$, then $X$—$X^{\perp}$ is in
$E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ with $\deg(X^{\perp})=1$. Next,
we introduce some notation and terminology from [6] that will be used in this
paper. We will denote by
$\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$ the set of subspaces of
type $(m,2s+\gamma,s,\Gamma)$ of $\mathbb{F}_{q}^{(2\nu+\delta)}$ with respect
to $S$ and write
$N(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)=|\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)|.$
Let $P$ be a fixed subspace of type $(m,2s+\gamma,s,\Gamma)$ in
$\mathbb{F}_{q}^{(2\nu+\delta)}.$ We will denote by
$\mathcal{M}_{P}(m_{1},2s_{1}+\gamma_{1},s_{1},\Gamma_{1};m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$
the set of subspaces of type $(m_{1},2s_{1}+\gamma_{1},s_{1},\Gamma_{1})$
contained in $P$ and write
$N(m_{1},2s_{1}+\gamma_{1},s_{1},\Gamma_{1};m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)=|\mathcal{M}_{P}(m_{1},2s_{1}+\gamma_{1},s_{1},\Gamma_{1};m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)|.$
Then

$V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}=\mathcal{M}(1,0,0;2\nu+\delta,\Delta)\bigcup\mathcal{M}(1,1,0,1;2\nu+\delta,\Delta)\bigcup\mathcal{M}(1,1,0,z;2\nu+\delta,\Delta)$

and
$|V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}|=\frac{q^{2\nu+\delta}-1}{q-1}$
by [6, Theorem 6.26].

######  Theorem 2.2.

Let $\mathbb{F}_{q_{1}}$ and $\mathbb{F}_{q_{2}}$ be finite fields. Then
$Oi\big{(}2\nu_{1}+\delta_{1},q_{1}\big{)}\cong
Oi\big{(}2\nu_{2}+\delta_{2},q_{2}\big{)}$ if and only if $\nu_{1}=\nu_{2}$,
$\delta_{1}=\delta_{2}$ and $q_{1}=q_{2}$.

###### Proof.

First suppose that $\nu_{1}=\nu_{2}$, $\delta_{1}=\delta_{2}$ and
$q_{1}=q_{2}$. Then $Oi\big{(}2\nu_{1}+\delta_{1},q_{1}\big{)}\cong
Oi\big{(}2\nu_{2}+\delta_{2},q_{2}\big{)}$, completing the proof in one
direction.

For the other direction, suppose that
$Oi\big{(}2\nu_{1}+\delta_{1},q_{1}\big{)}\cong
Oi\big{(}2\nu_{2}+\delta_{2},q_{2}\big{)}$. We need only consider
$Oi\big{(}2\nu_{i}+\delta_{i},q_{i}\big{)}$ for $i=1,2$. Then
$(q_{1}^{2\nu_{1}+\delta_{1}}-1)/(q_{1}-1)=(q_{2}^{2\nu_{2}+\delta_{2}}-1)/(q_{2}-1)$.
Take $M_{i}$ being a maximum complete subgraph (without considering circles)
of $Oi\big{(}2\nu_{i}+\delta_{i},1,q_{i}\big{)}$ such that
$XS_{2\nu_{i}+\delta_{i}}\,^{t}\\!Y=0$ for all different $X,Y\in M_{i}$. Let
$N_{i}=\\{Z\in M_{i}:ZS_{2\nu_{i}+\delta_{i}}\,^{t}\\!Z\neq 0\\}$ for $i=1,2$.
Then $\nu_{1}+\delta_{1}=|M_{1}|=|M_{2}|=\nu_{2}+\delta_{2}\ {\rm and}\
\delta_{1}=|N_{1}|=|N_{2}|=\delta_{2}.$ Thus $\nu_{1}=\nu_{2}$,
$\delta_{1}=\delta_{2}$ and $q_{1}=q_{2}$.

######  Proposition 2.3.

Let $Oi\big{(}2\nu+\delta,q\big{)}$ be the orthogonal inner graph over
$\mathbb{F}_{q}$ and $\sigma\in{\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$. Then
$\sigma\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}=Oi\big{(}2\nu+\delta,1,q\big{)}$.

###### Proof.

Let $X\in V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}$ and $\sigma\in{\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$. Then there exists $A\in
V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ with
$XS_{2\nu+\delta}\,^{t}\\!A=0$ and deg$(A)=2\nu+\delta-1$. Since
$\sigma\in{\rm Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$, it follows
that $\sigma(X)S_{2\nu+\delta}\,^{t}\\!(\sigma(A))=0$ and
deg$(\sigma(A))=2\nu+\delta-1$. So $\sigma(X)\in
V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}$. Since $\sigma$ is injective
and $V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}$ is finite, we have
$\sigma\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}=Oi\big{(}2\nu+\delta,1,q\big{)}$.

It follows that the complement graph of the orthogonal graph in [3] is a
proper subgraph of $Oi\big{(}2\nu+\delta,1,q\big{)}$. In fact, the vertices
set of the orthogonal graph in [3] contains all $1$-dimensional totally
isotropic orthogonal subspaces of $\mathbb{F}_{q}^{(2\nu+\delta)}$. But
$Oi\big{(}2\nu+\delta,1,q\big{)}$ contains all $1$-dimensional orthogonal
subspaces of $\mathbb{F}_{q}^{(2\nu+\delta)}$.

######  Proposition 2.4.

Let $T\in O_{2\nu+\delta}\big{(}\mathbb{F}_{q}\big{)}$ and

$\sigma_{T}:V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}\longrightarrow
V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)},A\longmapsto AT,$

for $\delta=0,1$ or $2.$ Then

${\rm(1)}$ $\sigma_{T}\in{\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)},$

${\rm(2)}$ For any $T_{1},T_{2}\in
O_{2\nu+\delta}\big{(}\mathbb{F}_{q}\big{)}$, $\sigma_{T_{1}}=\sigma_{T_{2}}$
if and only if $T_{1}=\pm T_{2}$.

###### Proof.

(1) Let $T\in O_{2\nu+\delta}\big{(}\mathbb{F}_{q}\big{)}$. Then $T$ is
nonsingular and $\sigma_{T}$ is a bijection. For any $A,B\in
V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$, we know that
$AS\,^{t}\\!B=ATS\,^{t}\\!(AT)$, then $A\text{---}B$ if and only if
$\sigma_{T}(A)\text{---}\sigma_{T}(B)$. Thus we conclude that
$\sigma_{T}\in{\rm Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$.

(2) First of all, if $T_{1}=\pm T_{2}\in
O_{2\nu+\delta}\big{(}\mathbb{F}_{q}\big{)}$, then it follows that
$\sigma_{T_{1}}=\sigma_{T_{2}}$. Conversely, suppose that
$\sigma_{T_{1}}=\sigma_{T_{2}}$. Then we need only consider them acting on
$Oi\big{(}2\nu+\delta,q\big{)}$. According to Proposition 2.3, we deduce that
$\sigma_{T_{i}}\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}=Oi\big{(}2\nu+\delta,1,q\big{)}$
for $i=1,2$. The following proof is similar to that of [3, Proposition 3.1].

Note that every orthogonal matrix of $O_{2\nu+\delta}(\mathbb{F}_{q})$ can
induce an automorphism of $Oi\big{(}2\nu+\delta,q\big{)}$ and two different
orthogonal matrices $T_{1}$ and $T_{2}$ induce the same automorphism of
$Oi\big{(}2\nu+\delta,q\big{)}$ if and only if $T_{1}=\pm T_{2}$. Thus
$PO_{2\nu+\delta}(\mathbb{F}_{q})$ can be regarded as a subgroup of ${\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$.

######  Proposition 2.5.

Let $Oi\big{(}2\nu+\delta,1,q\big{)}$ be an induced subgraph of
$Oi\big{(}2\nu+\delta,q\big{)}$ and $\sigma\in{\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$. Then $X$ and $\sigma(X)$ are
the same type orthogonal subspace for $X\in
V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}$.

###### Proof.

In order to prove the theorem, we need only consider three cases: $\delta=0,1$
and $2$. Without loss of generality, we need only consider the case of
$\delta=0$. Other cases are similar to prove. Let $\delta=0$.

Let $X\in V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}$. Then the type of
$X$ is one of $(1,0,0)$, $(1,1,0,1)$ and $(1,1,0,z)$. If the type of $X$ is
$(1,0,0)$, then it is easy to check that the type of $\sigma(X)$ is also
$(1,0,0)$. By [3, Theorem 6.4] and Proposition 2.4, we know that the sets
$\mathcal{M}(1,0,0;2\nu+\delta,\Delta)$,
$\mathcal{M}(1,1,0,1;2\nu+\delta,\Delta)$ and
$\mathcal{M}(1,1,0,z;2\nu+\delta,\Delta)$ are three orbits of
$Oi\big{(}2\nu+\delta,q\big{)}$ under the action of
$PO_{2\nu+\delta,\Delta}(\mathbb{F}_{q})$.

If the type of $X$ is $(1,1,0,1)$, then by [3, Corollary 6.6] we know that the
type of $X^{\perp}$ is

$\left\\{\begin{array}[]{ll}(2\nu-1,2\nu-1,\nu-1,1),&\text{if}~{}-1\in\mathbb{F}_{q}^{*2},\\\
(2\nu-1,2\nu-1,\nu-1,z),&\text{if}~{}-1\notin\mathbb{F}_{q}^{*2}.\end{array}\right.$

If the type of $X$ is $(1,1,0,z)$, then by [3, Corollary 6.6] we know that the
type of $X^{\perp}$ is

$\left\\{\begin{array}[]{ll}(2\nu-1,2\nu-1,\nu-1,z),&\text{if}~{}-1\in\mathbb{F}_{q}^{*2},\\\
(2\nu-1,2\nu-1,\nu-1,1),&\text{if}~{}-1\notin\mathbb{F}_{q}^{*2}.\end{array}\right.$

Let

$\mathcal{X}=\\{Y\in
V\big{(}Oi\big{(}2\nu,q\big{)}\big{)}:YS_{2\nu}{{}^{t}\\!X}=0{\rm\ and\ the\
type\ of\ Y\ is\ (1,1,0,1)}\\}.$

If the type of $X$ is $(1,1,0,1)$, then by [3, Theorem 6.33], we have

$|\mathcal{X}|=\left\\{\begin{array}[]{ll}N(1,1,0,1;2\nu-1,2\nu-1,\nu-1,1;2\nu),&\text{if}~{}-1\in\mathbb{F}_{q}^{*2},\\\
N(1,1,0,1;2\nu-1,2\nu-1,\nu-1,z;2\nu),&\text{if}~{}-1\notin\mathbb{F}_{q}^{*2},\end{array}\right.$
$=\left\\{\begin{array}[]{ll}N(1,1,0,1;2\nu-1,1),&\text{if}~{}-1\in\mathbb{F}_{q}^{*2},\\\
N(1,1,0,1;2\nu-1,z),&\text{if}~{}-1\notin\mathbb{F}_{q}^{*2}.\end{array}\right.$

If the type of $X$ is $(1,1,0,z)$, then by [3, Theorem 6.33], we have

$|\mathcal{X}|=\left\\{\begin{array}[]{ll}N(1,1,0,1;2\nu-1,2\nu-1,\nu-1,z;2\nu),&\text{if}~{}-1\in\mathbb{F}_{q}^{*2},\\\
N(1,1,0,1;2\nu-1,2\nu-1,\nu-1,1;2\nu),&\text{if}~{}-1\notin\mathbb{F}_{q}^{*2},\end{array}\right.$
$=\left\\{\begin{array}[]{ll}N(1,1,0,1;2\nu-1,z),&\text{if}~{}-1\in\mathbb{F}_{q}^{*2},\\\
N(1,1,0,1;2\nu-1,1),&\text{if}~{}-1\notin\mathbb{F}_{q}^{*2}.\end{array}\right.$

If the type of $X_{1}$ is $(1,1,0,1)$ and the type of $X_{2}$ is $(1,1,0,z)$,
then by [3, Theorem 6.26], we have $|\mathcal{X}_{1}|\neq|\mathcal{X}_{2}|$.
Thus by Proposition 2.3, we know that $X$ and $\sigma(X)$ are the same type
orthogonal subspace for $X\in V\big{(}Oi\big{(}2\nu+\delta,1,q\big{)}\big{)}$.

##  3 Automorphism groups of orthogonal inner graphs

In order to determine automorphism groups ${\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ of orthogonal inner graphs
$Oi(2\nu+\delta,\mathbb{F}_{q})$, we need only consider three cases of
$\delta=0,1$ and $2$.

Firstly, we will discuss the case of $\delta=0$. Let $S=S_{2\nu}$. In what
follows, we write $f_{i}$ for $e_{\nu+i},$ $1\leq i\leq\nu$. Then

$e_{i}{S_{2\nu}}{{}^{t}\\!f_{i}}=1,e_{i}{S_{2\nu}}{{}^{t}\\!e_{j}}=f_{i}{S_{2\nu}}{{}^{t}\\!f_{j}}=0\
{\rm for}\ 1\leq i,j\leq\nu$

and

$e_{i}{S_{2\nu}}{{}^{t}\\!f_{j}}=0\ {\rm for}\ i\neq j,\ 1\leq i,j\leq\nu.$

Let $\varphi_{2\nu}$ be the natural action of ${\rm Aut}(\mathbb{F}_{q})$ on
the group
$\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}$
defined by

$\displaystyle\varphi_{2\nu}:\big{(}\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\big{)}\times{\rm
Aut}(\mathbb{F}_{q})$ $\displaystyle\longrightarrow$
$\displaystyle\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}$
$\displaystyle((k_{1},k_{2},\ldots,k_{\nu}),\pi)$
$\displaystyle\longrightarrow$
$\displaystyle(\pi(k_{1}),\pi(k_{2}),\ldots,\pi(k_{\nu}))$

for $\pi\in{\rm Aut}(\mathbb{F}_{q})$, $k_{1}\in\mathbb{F}_{q}^{*2}$ and
$k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$. The _semidirect product_ of
$\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}$
by ${\rm Aut}(\mathbb{F}_{q})$ with respect to $\varphi_{2\nu}$, denoted by
$(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*})\rtimes_{\varphi_{2\nu}}{\rm
Aut}(\mathbb{F}_{q})$, is the group consisting of all elements of the form
$(k_{1},k_{2}\ldots,k_{\nu},\pi)$, with the multiplication defined by

$(k_{1},k_{2},\ldots,k_{\nu},\pi)(k_{1}^{{}^{\prime}},k_{2}^{{}^{\prime}},\ldots,k_{\nu}^{{}^{\prime}},\pi^{{}^{\prime}})=(k_{1}\pi(k_{1}^{{}^{\prime}}),k_{2}\pi(k_{2}^{{}^{\prime}}),\ldots,k_{\nu}\pi(k_{\nu}^{{}^{\prime}}),\pi\pi^{{}^{\prime}}).$

Define maps:

$\displaystyle\sigma_{\pi}:V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$
$\displaystyle\longrightarrow$ $\displaystyle
V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$
$\displaystyle\big{(}a_{ij}\big{)}_{m\times(2\nu+\delta)}$
$\displaystyle\longrightarrow$
$\displaystyle\big{(}\pi(a_{ij})\big{)}_{m\times(2\nu+\delta)}$

and

$\displaystyle\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}:V\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$
$\displaystyle\longrightarrow$ $\displaystyle
V\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$ $\displaystyle A$
$\displaystyle\longrightarrow$ $\displaystyle\sigma_{\pi}(A)\cdot{\rm
diag}(k_{1},k_{2},\ldots,k_{\nu},k_{1}^{-1},k_{2}^{-1},\ldots,k_{\nu}^{-1}),$

where $\pi\in{\rm Aut}(\mathbb{F}_{q})$, $k_{1}\in\mathbb{F}_{q}^{*2}$ and
$k_{2}\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$. It is easy to check that
$\sigma_{\pi}\in{\rm Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$. Moreover,
since ${\rm
diag}(k_{1},k_{2},\ldots,k_{\nu},k_{1}^{-1},k_{2}^{-1}\ldots,k_{\nu}^{-1})S_{2\nu}\,{{}^{t}({\rm
diag}(k_{1},k_{2},\ldots,k_{\nu},k_{1}^{-1},k_{2}^{-1},\ldots,k_{\nu}^{-1}))}=S_{2\nu},$
we have ${\rm
diag}(k_{1},k_{2},\ldots,k_{\nu},k_{1}^{-1},\ldots,k_{\nu}^{-1})\in
O_{2\nu}(\mathbb{F}_{q}).$ So
$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}\in{\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}.$

Before considering the problem of automorphism groups of
$Oi\big{(}2\nu,q\big{)}$, let′s look at a simple fact.

######  Example 3.1.

Let $\mathbb{F}_{q}$ be a field. Then

$V\big{(}Oi\big{(}2,q\big{)}\big{)}=\\{[(1,x)]:x\in\mathbb{F}_{q}\\}\bigcup\\{[(0,1)]\\}$

and

$E\big{(}Oi\big{(}2,q\big{)}\big{)}=\\{[(1,x)]\text{---}[(1,y)]:xy=-1,x,y\in\mathbb{F}_{q}\\}\bigcup\\{[(1,0)]\text{---}[(1,0)],[(0,1)]\text{---}[(0,1)]\\}.$

Clearly, $Oi\big{(}2,q\big{)}$ is a disconnected graph.

######  Theorem 3.2.

Let $\nu\geq 2$ and $E_{2\nu}$ be the subgroup of ${\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$ defined as follows:

$E_{2\nu}=\big{\\{}\sigma\in{\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}:\sigma\big{(}[e_{i}]\big{)}=[e_{i}]~{}and~{}\sigma\big{(}[f_{i}]\big{)}=[f_{i}]~{}\text{for}~{}1\leq
i\leq\nu\big{\\}}.$

Then ${\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}=PO_{2\nu}\big{(}\mathbb{F}_{q}\big{)}\cdot
E_{2\nu}$. Moreover, let

$\displaystyle
h:\left.{\big{(}(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*})\rtimes_{\varphi_{2\nu}}{\rm
Aut}(\mathbb{F}_{q})\big{)}}\middle/K_{2\nu}\right.$
$\displaystyle\longrightarrow$ $\displaystyle E_{2\nu}$
$\displaystyle(k_{1},k_{2},\ldots,k_{\nu},\pi)K_{2\nu}$
$\displaystyle\longrightarrow$
$\displaystyle\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)},$

where $1_{\mathbb{F}_{q}}$ is an identity element of ${\rm
Aut}(\mathbb{F}_{q})$ and

$K_{2\nu}=\left\\{\begin{array}[]{ll}\\{(c,c,\ldots,c,1_{\mathbb{F}_{q}}):c=\pm
1\\},&\ {\rm if}\ -1\in\mathbb{F}_{q}^{*2},\\\
\\{(1,1,\ldots,1,1_{\mathbb{F}_{q}})\\},&\ {\rm if}\
-1\notin\mathbb{F}_{q}^{*2}.\end{array}\right.$

Then $h$ is a group isomorphism from
$\left.{\big{(}(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*})\rtimes_{\varphi_{2\nu}}{\rm
Aut}(\mathbb{F}_{q})\big{)}}\middle/K_{2\nu}\right.$ to $E_{2\nu}$.

###### Proof.

Let $\tau\in{\rm Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$. By Proposition
2.3, we can suppose that $\tau\big{(}[e_{i}]\big{)}=[\alpha_{i}]$ and
$\tau\big{(}[f_{i}]\big{)}=[\beta_{i}]$ for $1\leq i\leq\nu$. Then

$\alpha_{i}{S_{2\nu}}{{}^{t}\\!\\!\beta}_{i}\neq
0,\alpha_{i}S_{2\nu}{{}^{t}\\!\\!\alpha}_{j}=\beta_{i}S_{2\nu}{{}^{t}\\!\\!\beta}_{j}=0\
{\rm for}\ 1\leq i,j\leq\nu$

and

$\alpha_{i}S_{2\nu}{{}^{t}\\!\\!\beta}_{j}=0\ {\rm for}\ i\neq j,\,1\leq
i,j\leq\nu.$

Let $A=[e_{1},\ldots,e_{\nu},f_{1},\ldots,f_{\nu}]$ and
$A^{{}^{\prime}}=[\alpha_{1},\ldots,\alpha_{\nu},\beta_{1},\ldots,\beta_{\nu}]$.
Since $A^{{}^{\prime}}S\,^{t}\\!A^{{}^{\prime}}$ is a $2\nu\times 2\nu$
nonsingular symmetric matrix, we can choose $\alpha_{i},\beta_{i},1\leq
i\leq\nu,$ such that $\alpha_{i}S\,^{t}\\!\beta_{i}=1$. Then
$AS\,^{t}\\!A=A^{{}^{\prime}}S\,^{t}\\!A^{{}^{\prime}}$. By [6, Lemma 6.8],
there exists $T\in O_{2\nu}(\mathbb{F}_{q})$ such that $A=A^{{}^{\prime}}T$.
Let $\tau_{1}=\sigma_{T}\tau$. Then $\tau_{1}([e_{i}])=[e_{i}]$ and
$\tau_{1}([f_{i}])=[f_{i}]$ for $1\leq i\leq\nu$. So $\tau_{1}\in E_{2\nu}$
and $\tau=\sigma_{T}^{-1}\tau_{1}\in PO_{2\nu}(\mathbb{F}_{q})\cdot E_{2\nu}$.
Thus ${\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}=PO_{2\nu}(\mathbb{F}_{q})\cdot
E_{2\nu}$.

Let $k_{1}\in\mathbb{F}_{q}^{*2},~{}k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$
and $\pi\in{\rm Aut}(\mathbb{F}_{q})$. Then it is easy to check that

$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}([e_{i}])=[e_{i}]\ {\rm and}\
\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}([f_{i}])=[f_{i}]$

for $1\leq i\leq\nu,$ and so $\sigma_{(k_{1},k_{2}\ldots,k_{\nu},\pi)}\in
E_{2\nu}$. Let

$\displaystyle
h^{{}^{\prime}}:(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*})\rtimes_{\varphi_{2\nu}}{\rm
Aut}(\mathbb{F}_{q})$ $\displaystyle\longrightarrow$ $\displaystyle E_{2\nu}$
$\displaystyle(k_{1},k_{2},\ldots,k_{\nu},\pi)$ $\displaystyle\longrightarrow$
$\displaystyle\sigma_{(k_{1},\ldots,k_{\nu},\pi)}.$

Then by the definition of $\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}$ and
Proposition2.4, it is easily seen that $h^{{}^{\prime}}$ is a group
homomorphism and the kernel of $h^{{}^{\prime}}$ is $K_{2\nu}$. Hence in order
to show that $h$ is a group isomorphism from
$\left.{\big{(}(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*})\rtimes_{\varphi_{2\nu}}{\rm
Aut}(\mathbb{F}_{q})\big{)}}\middle/K_{2\nu}\right.$ to $E_{2\nu}$, it
suffices to show that for any $\sigma\in E_{2\nu+1}$, there exist
$k_{1}\in\mathbb{F}_{q}^{*2},$ $k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$ and
$\pi\in{\rm Aut}(\mathbb{F}_{q})$ such that
$\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}.$

Let $\sigma\in E_{2\nu}$. And then we will prove that there exist
$k_{1}\in\mathbb{F}_{q}^{*2},$ $k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$ and
$\pi\in{\rm Aut}(\mathbb{F}_{q})$ such that

$\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}.$

Let $[(a_{1},\ldots,a_{2\nu})]\in V\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$ and
suppose that $\sigma([(a_{1},\ldots,a_{2\nu})])=[(b_{1},\ldots,b_{2\nu})]$.
Then

$a_{i}=0$ if and only if $[(a_{1},\ldots,a_{2\nu})]$—$[f_{i}]$

and

$a_{\nu+i}=0$ if and only if $[(a_{1},\ldots,a_{2\nu})]$—$[e_{i}]$

for $1\leq i\leq\nu$. Since $\sigma([e_{i}])=[e_{i}]$ and
$\sigma([f_{i}])=[f_{i}]$, $1\leq i\leq\nu$, we have

$a_{k}=0\text{ if and only if }b_{k}=0\text{ for }1\leq k\leq 2\nu.$

Moreover, $[(a_{1},\ldots,a_{2\nu})]\in V\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$
can be written uniquely as
$[(0,\ldots,0,1,a_{i+1}^{{}^{\prime}},\ldots,a_{2\nu}^{{}^{\prime}})]$ if
$a_{1}=\cdots=a_{i-1}=0$ and $a_{i}\neq 0$. Therefore
$\sigma([(0,\ldots,0,1,a_{i+1}^{{}^{\prime}},\ldots,a_{2\nu}^{{}^{\prime}})])$
can be written uniquely as $[(0,\ldots,0,1,b_{i+1},\ldots,b_{2\nu})].$

Let $1\leq k\leq\nu$. For $[f_{1}+ae_{k}],[f_{1}+bf_{k}]\in
V\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$, there exist
$a^{{}^{\prime}},b^{{}^{\prime}}\in\mathbb{F}_{q}$ such that
$\sigma\big{(}[e_{1}+ae_{k}]\big{)}=[e_{1}+a^{{}^{\prime}}e_{k}]$ and
$\sigma\big{(}[e_{1}+bf_{k}]\big{)}=[e_{1}+b^{{}^{\prime}}e_{k}]$. Thus we can
define permutations $\pi_{i}$ of $\mathbb{F}_{q}$ with $\pi_{i}(0)=0$ such
that

$\sigma\big{(}[e_{1}+a_{i}e_{i}]\big{)}=[e_{1}+\pi_{i}(a_{i})e_{i}]\,\ {\rm
and}\,\
\sigma\big{(}[e_{1}+a_{\nu+i}f_{i}]\big{)}=[e_{1}+\pi_{\nu+i}(a_{\nu+i})f_{i}],$

where $2\leq i\leq\nu.$

In the following proof, we need only consider two cases: $\nu=2$ and $\nu>2$.

(1) $\nu=2$. Clearly, $V\big{(}Oi\big{(}4,q\big{)}\big{)}$ consists of four
types of vertices:
$\\{[(1,a_{2},a_{3},a_{4})]:a_{2},a_{3},a_{4}\in\mathbb{F}_{q}\\},\\{[(0,1,\linebreak[1]a_{3},a_{4})]:a_{3},a_{4}\in\mathbb{F}_{q}\\},\\{[(0,0,1,a_{4})]:a_{4}\in\mathbb{F}_{q}\\}$
and $\\{[(0,0,0,1)]\\}$. If
$\sigma([(x_{1},x_{2},x_{3},x_{4})])=[(y_{1},y_{2},y_{3},\linebreak[1]y_{4})]$
and $x_{i}=0$, then we get $y_{i}=0$ for $1\leq i\leq 4$. Thus without
restriction of generality we can assume that $a_{i}\neq 0$ for $1\leq i\leq
4$. Now we first determine that
$\sigma([(0,0,1,a_{4})]),\sigma([(0,1,a_{3},a_{4})])$ and
$\sigma([(1,a_{2},a_{3},a_{4})])$ for
$a_{2},a_{3},\linebreak[1]a_{4}\in\mathbb{F}_{q}$.

Suppose that $\sigma([(0,0,1,a_{4})])=[(0,0,1,a_{4}^{{}^{\prime}})]$. Since
$[(0,0,1,a_{4})]\text{---}[(1,-a_{4}^{-1},0,0)]$,
$[(0,0,1,a_{4}^{{}^{\prime}})]\text{---}\allowbreak[(1,\pi_{2}(-a_{4}^{-1}),0,0)]$,
i.e., $a_{4}^{{}^{\prime}}=-\pi_{2}(-a_{4}^{-1})^{-1}$. Thus
$\sigma([(0,0,1,a_{4})])=[(0,0,1,-\pi_{2}(-a_{4}^{-1})^{-1})]$.

Suppose that
$\sigma([(0,1,a_{3},a_{4})])=[(0,1,a_{3}^{{}^{\prime}},a_{4}^{{}^{\prime}})]$.
Since $[(0,1,a_{3},a_{4})]\text{---}[(1,0,0,-a_{3})]$, it follows that
$[(0,1,a_{3}^{{}^{\prime}},a_{4}^{{}^{\prime}})]\text{---}[(1,0,0,\pi_{4}(-a_{3}))]$,
i.e., $a_{3}^{{}^{\prime}}=-\pi_{4}(-a_{3})$. Since
$[(0,1,a_{3},a_{4})]\text{---}[(1,-a_{3}a_{4}^{-1},0,\linebreak[1]0)]$, we see
that
$[(0,1,-\pi_{4}(-a_{3}),a_{4}^{{}^{\prime}})]\text{---}[(1,\pi_{2}(-a_{3}a_{4}^{-1}),0,0)]$,
i.e., $a_{4}^{{}^{\prime}}=\pi_{2}(-a_{3}a_{4}^{-1})^{-1}\pi_{4}(-a_{3})$.
Thus
$\sigma([(0,1,a_{3},a_{4})])=[(0,1,-\pi_{4}(-a_{3}),\pi_{2}(-a_{3}a_{4}^{-1})^{-1}\pi_{4}(-a_{3}))].$

Similarly, we can prove that

$\sigma([(1,a_{2},\linebreak[1]a_{3},a_{4})])=[(1,\pi_{2}(a_{2}),-\pi_{2}(-a_{3}a_{4}^{-1})\pi_{4}(a_{4}),\pi_{4}(a_{4}))].$

Next we will prove that $\pi$ is an automorphism of $\mathbb{F}_{q}$ for
$\pi=\pi_{2}(1)^{-1}\pi_{2}=\pi_{4}(1)^{-1}\pi_{4}$. Let
$a,b\in\mathbb{F}_{q}$.

Since $[(1,a,0,a)]\text{---}[(1,1,0,-1)]$ and
$[(1,a,0,a)]\text{---}[(1,-1,0,1)]$, we have
$[(1,\pi_{2}(a),0,\pi_{4}(a))]\text{---}\linebreak[1][(1,\pi_{2}(1),0,\pi_{4}(-1))]$
and $[(1,\pi_{2}(a),0,\pi_{4}(a))]\text{---}[(1,\pi_{2}(-1),0,\pi_{4}(1))]$,
i.e., $\pi_{2}(a)\pi_{4}(-1)+\pi_{2}(1)\pi_{4}(a)=0$ and
$\pi_{2}(a)\pi_{4}(1)+\pi_{2}(-1)\pi_{4}(a)=0$. Let $a=1$, then
$-\pi_{2}(1)=\pi_{2}(-1)$ and $-\pi_{4}(1)=\pi_{4}(-1)$. Thus we deduce that
$\pi_{2}(1)^{-1}\pi_{2}=\pi_{4}(1)^{-1}\pi_{4}$. Set
$\pi=\pi_{2}(1)^{-1}\pi_{2}=\pi_{4}(1)^{-1}\pi_{4}$.

Since $[(1,-a,0,a)]\text{---}[(1,1,0,1)]$, we have
$[(1,\pi_{2}(-a),0,\pi_{4}(a))]\text{---}[(1,\pi_{2}(1),0,\linebreak[1]\pi_{4}(1))]$,
i.e., $\pi_{2}(-a)\pi_{4}(1)+\pi_{2}(1)\pi_{4}(a)=0$. It follows that
$\pi_{2}(-a)=-\pi_{2}(a)$ since
$\pi_{2}(1)^{-1}\pi_{2}(-a)=-\pi_{4}(1)^{-1}\pi_{4}(a)=-\pi_{2}(1)^{-1}\pi_{2}(a)$.
Similarly, we have $\pi_{4}(-a)=-\pi_{4}(a)$.

Since $[(1,ab,0,a)]\text{---}[(1,b,0,-1)]$, we have
$[(1,\pi_{2}(ab),0,\pi_{4}(a))]\text{---}[(1,\pi_{2}(b),0,-\pi_{4}(1))]$,
i.e., $-\pi_{2}(ab)\pi_{4}(1)+\pi_{4}(a)\pi_{2}(b)=0$. Then
$\pi_{2}(ab)=\pi_{4}(1)^{-1}\pi_{4}(a)\pi_{2}(b)=\pi_{2}(1)^{-1}\pi_{2}(a)\pi_{2}(b).$
Thus

$\pi(ab)=\pi_{2}(1)^{-1}\pi_{2}(ab)=\pi_{2}(1)^{-1}\pi_{2}(1)^{-1}\pi_{2}(a)\pi_{2}(b)=\pi(a)\pi(b).$

Let $ab\neq 0$. Since $[(1,a+b,0,1)]\text{---}[(0,1,ab^{-1},-b^{-1})]$, we
have
$[(1,\pi_{2}(a+b),0,\pi_{4}(1))]\text{---}[(0,1,\linebreak[1]\pi_{4}(ab^{-1}),-\pi_{4}(ab^{-1})\pi_{2}(a)^{-1})]$,
i.e.,
$\pi_{4}(ab^{-1})-\pi_{2}(a+b)\pi_{4}(ab^{-1})\pi_{2}(a)^{-1}+\pi_{4}(1)=0$.
Then

$\pi_{2}(a+b)=\pi_{2}(a)+\pi_{2}(a)\pi_{4}(1)\pi_{4}(ab^{-1})^{-1}=\pi_{2}(a)+\pi_{2}(1)\pi(a)\pi(ab^{-1})^{-1}~{}~{}$
$~{}\,=\pi_{2}(a)+\pi_{2}(1)\pi(a)\pi(a^{-1}b)=\pi_{2}(a)+\pi_{2}(b).~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\>~{}\>\,$

Thus
$\pi(a+b)=\pi_{2}(1)^{-1}\pi_{2}(a+b)=\pi_{2}(1)^{-1}(\pi_{2}(a)+\pi_{2}(b))=\pi(a)+\pi(b).$

Since $\pi_{2}$ is a permutation of $\mathbb{F}_{q}$, it follows that $\pi$ is
injective. $\pi$ is surjective since $\mathbb{F}_{q}$ is finite. Thus $\pi$ is
an automorphism of $\mathbb{F}_{q}$. Furthermore, we conclude that

$\sigma([(a_{1},a_{2},a_{3},a_{4})])=[(\pi(a_{1}),\pi_{2}(1)\pi(a_{2}),\pi_{2}(1)\pi_{4}(1)\pi(a_{3}),\pi_{4}(1)\pi(a_{4}))].$

Lastly, we will prove that there exist
$k_{1}\in\mathbb{F}_{q}^{*2},k_{2}\in\mathbb{F}_{q}^{*}$ and $\pi\in{\rm
Aut}(\mathbb{F}_{q})$ such that $\sigma=\sigma_{(k_{1},k_{2},\pi)}$.

By Proposition 2.5, we know that $[(1,0,1,0)]$ and $\sigma([(1,0,1,0)])$ are
the same type orthogonal subspace. Then we have
$\pi_{2}(1)\pi_{4}(1)\in\mathbb{F}_{q}^{*2}$. Let
$k_{1}=\pi_{2}(1)\pi_{4}(1),$ $k_{2}=\pi_{2}(1)$ and
$\pi=\pi_{2}(1)^{-1}\pi_{2}=\pi_{4}(1)^{-1}\pi_{4}$. Then for
$\big{(}a_{ij}\big{)}_{m\times 4}\in V\big{(}Oi\big{(}4,q\big{)}\big{)}$, we
have

$\sigma\big{(}\big{(}a_{ij}\big{)}_{m\times
4}\big{)}=\left(\begin{array}[]{cccc}k_{1}\pi(a_{11})&k_{2}\pi(a_{12})&k_{1}^{-1}\pi(a_{13})&k_{2}^{-1}\pi(a_{14})\\\
\vdots&\vdots&\vdots&\vdots\\\
k_{1}\pi(a_{m1})&k_{2}\pi(a_{m2})&k_{1}^{-1}\pi(a_{m3})&k_{2}^{-1}\pi(a_{m4})\end{array}\right).$

So $\sigma\big{(}\big{(}a_{ij}\big{)}_{m\times
4}\big{)}=\sigma_{(k_{1},k_{2},\pi)}\big{(}\big{(}a_{ij}\big{)}_{m\times
4}\big{)}$ and therefore $\sigma=\sigma_{(k_{1},k_{2},\pi)}.$

(2) $\nu>2$. Similar to the case of $\nu=2$, we can prove that there exist
$k_{1}\in\mathbb{F}_{q}^{*2},k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$ and
$\pi\in{\rm Aut}(\mathbb{F}_{q})$ such that
$\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}.$

######  Corollary 3.3.

Let $\nu\geq 1$ and $\mathbb{F}_{q}$ be a field of characteristic $p$. Then

$\big{|}{\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}\big{|}=\left\\{\begin{array}[]{ll}2^{\frac{q+1}{2}}\cdot(\frac{q-1}{2})!,&{\rm\
if}\ \nu=1;\\\
\frac{1}{2}q^{\nu(\nu-1)}\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}-1)\mathop{\prod}\limits_{i=1}^{\nu-1}(q^{i}+1)[\mathbb{F}_{q}:\mathbb{F}_{p}],&\
{\rm if}\ \nu\geq 2\ {\rm and\ }-1\in\mathbb{F}_{q}^{*2};\\\
q^{\nu(\nu-1)}\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}-1)\mathop{\prod}\limits_{i=1}^{\nu-1}(q^{i}+1)[\mathbb{F}_{q}:\mathbb{F}_{p}],&\
{\rm if}\ \nu\geq 2\ {\rm and\
}-1\notin\mathbb{F}_{q}^{*2}.\end{array}\right.$

###### Proof.

(1) Let $\nu=1$. Vertices $[e_{1}],[e_{2}]$ in $Oi\big{(}2,q\big{)}$ are
totally isotropic subspaces. According to Example 3.1, it is easy to calculate
that $\big{|}{\rm
Aut}\big{(}Oi\big{(}2,q\big{)}\big{)}\big{|}=2^{\frac{q+1}{2}}\cdot(\frac{q-1}{2})!$.

(2) Let $\nu\geq 2$ and $-1\in\mathbb{F}_{q}^{*2}$. Then $\big{|}{\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}\big{|}=\frac{|PO_{2\nu}(\mathbb{F}_{q})|\cdot|E_{2\nu}|}{|PO_{2\nu}(\mathbb{F}_{q})\cap
E{2\nu}|}.$ According to [6, Theorem 6.21], we see that
$|PO_{2\nu}(\mathbb{F}_{q})|=\frac{1}{2}|O_{2\nu}(\mathbb{F}_{q})|=\frac{1}{2}q^{\nu(\nu-1)}\prod_{i=1}^{\nu}(q^{i}-1)\prod_{i=0}^{\nu-1}(q^{i}+1).$
Clearly, we have
$|E_{2\nu}|=\frac{1}{4}(q-1)^{\nu}\cdot[\mathbb{F}_{q}:\mathbb{F}_{p}]$, where
$|{\rm Aut}(\mathbb{F}_{q})|=[\mathbb{F}_{q}:\mathbb{F}_{p}]$. Let
$\sigma_{P}\in PO_{2\nu}(\mathbb{F}_{q})\cap E_{2\nu}$. Then
$\sigma_{P}=\sigma_{(k_{1},\ldots,k_{\nu},\pi)}$ for some
$k_{1}\in\mathbb{F}_{q}^{*2},k_{1},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$ and
$\pi\in{\rm Aut}(\mathbb{F}_{q})$. So $|PO_{2\nu}(\mathbb{F}_{q})\cap
E_{2\nu}|=\frac{1}{4}(q-1)^{\nu}$ and thus

$~{}~{}\big{|}{\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}\big{|}=\frac{1}{2}q^{\nu(\nu-1)}\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}-1)\mathop{\prod}\limits_{i=1}^{\nu-1}(q^{i}+1)[\mathbb{F}_{q}:\mathbb{F}_{p}].$

(3) Let $\nu\geq 2$ and $-1\notin\mathbb{F}_{q}^{*2}$. Similar to the case of
$(2)$, we can prove that $\big{|}{\rm
Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}\big{|}=q^{\nu(\nu-1)}\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}-1)\mathop{\prod}\limits_{i=1}^{\nu-1}(q^{i}+1)[\mathbb{F}_{q}:\mathbb{F}_{p}].$

Secondly, we will discuss the case of $\delta=1$. Since the groups
$O_{2\nu+1,1}(\mathbb{F}_{q})$ and $O_{2\nu+1,z}(\mathbb{F}_{q})$ are
isomorphic, we need only consider the case of $O_{2\nu+1,1}(\mathbb{F}_{q})$.
The other case is similar to prove. Let $S=S_{2\nu+1,1}$. In what follows, we
write $\varepsilon$ for $e_{2\nu+1}$. Then

$e_{i}{S_{2\nu+1,1}}{{}^{t}\\!f_{i}}=\varepsilon
S_{2\nu+1,1}{{}^{t}\\!\varepsilon}=1,e_{i}{S_{2\nu+1,1}}{{}^{t}\\!e_{j}}=f_{i}{S_{2\nu+1,1}}{{}^{t}\\!f_{j}}=0\
{\rm for}\ 1\leq i,j\leq\nu$

and

$e_{i}{S_{2\nu+1,1}}{{}^{t}\\!f_{j}}=e_{i}{S_{2\nu+1,1}}{{}^{t}\\!\varepsilon}=f_{i}{S_{2\nu+1,1}}{{}^{t}\\!\varepsilon}=0\
{\rm for}\ i\neq j,1\leq i,j\leq\nu.$

Similar to the case of $\delta=0$, let $\varphi_{2\nu+1}$ be the natural
action of ${\rm Aut}(\mathbb{F}_{q})$ on the group
$\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*}$
defined by
$\varphi_{2\nu+1}(\pi)((k_{1},k_{2},\ldots,k_{\nu},\delta_{1}))=(\pi(k_{1}),\pi(k_{2}),\ldots,\pi(k_{\nu}),\pi(\delta_{1}))$
and
$\big{(}\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*}\big{)}\rtimes_{\varphi_{2\nu+1}}{\rm
Aut}\big{(}\mathbb{F}_{q}\big{)}$ be a semidirect product group corresponding
to $\varphi_{2\nu+1}$. Moreover, we define maps
$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)}$ from
$V\big{(}Oi\big{(}2\nu+1,q\big{)}\big{)}$ to itself by

$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)}(A)=\sigma_{\pi}(A)\cdot{\rm
diag}(k_{1},k_{2},\ldots,k_{\nu},k_{1}^{-1},k_{2}^{-1},\ldots,k_{\nu}^{-1},\delta_{1}),$

where $k_{1}\in\mathbb{F}_{q}^{*2}$,
$k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$, $\delta_{1}\in\mathbb{F}_{3}^{*}$
and $\pi\in{\rm Aut}(\mathbb{F}_{q})$. It is easy to prove that
$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)}\in{\rm
Aut}\big{(}Oi\big{(}2\nu+1,q\big{)}\big{)}$.

######  Theorem 3.4.

Let $\nu\geq 2$ and $E_{2\nu+1}$ be the subgroup of ${\rm
Aut}\big{(}Oi\big{(}2\nu+1,q\big{)}\big{)}$ defined as follows

$\big{\\{}\sigma\in{\rm
Aut}\big{(}Oi\big{(}2\nu+1,q\big{)}\big{)}:\sigma\big{(}[e_{i}]\big{)}=[e_{i}],\sigma\big{(}[f_{i}]\big{)}=[f_{i}]\
{\rm and}\ \sigma\big{(}[\varepsilon]\big{)}=[\varepsilon]\ {\rm for}\ 1\leq
i\leq\nu\big{\\}}.$

Then ${\rm Aut}\big{(}Oi\big{(}2\nu+1,q\big{)}\big{)}=PO_{2\nu+1}(F_{q})\cdot
E_{2\nu+1}$. Moreover, let

$\displaystyle
h:\left.{\big{(}(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*})\rtimes_{\varphi_{2\nu+1}}{\rm
Aut}(\mathbb{F}_{q})\big{)}}\middle/K_{2\nu+1}\right.$
$\displaystyle\longrightarrow$ $\displaystyle E_{2\nu+1}$
$\displaystyle(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)K_{2\nu+1}$
$\displaystyle\longrightarrow$
$\displaystyle\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)},$

where $1_{\mathbb{F}_{q}}$ is an identity element of ${\rm
Aut}(\mathbb{F}_{q})$, and

$K_{2\nu+1}=\left\\{\begin{array}[]{ll}\\{(c,c,\ldots,c,c,1_{\mathbb{F}_{q}}):c=\pm
1\\},&\ {\rm if}\ -1\in\mathbb{F}_{q}^{*2},\\\
\\{(1,1,\ldots,1,1,1_{\mathbb{F}_{q}})\\},&\ {\rm if}\
-1\notin\mathbb{F}_{q}^{*2}.\end{array}\right.$

Then $h$ is an isomorphism of groups from
$\left.{\big{(}(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*})\rtimes_{\varphi_{2\nu+1}}{\rm
Aut}(\mathbb{F}_{q})\big{)}}\middle/K_{2\nu+1}\right.$ to $E_{2\nu+1}$.

###### Proof.

Similar to Theorem 3.2, we can prove that

${\rm
Aut}\big{(}Oi\big{(}2\nu+1,q\big{)}\big{)}=PO_{2\nu+1,1}(\mathbb{F}_{q})\cdot
E_{2\nu+1}.$

Define a map:

$\displaystyle
h^{{}^{\prime}}:(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*})\rtimes_{\varphi_{2\nu+1}}{\rm
Aut}(\mathbb{F}_{q})$ $\displaystyle\longrightarrow$ $\displaystyle
E_{2\nu+1}$ $\displaystyle(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)$
$\displaystyle\longrightarrow$
$\displaystyle\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)}.$

By the definition of $\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)}$,
it is easy to prove that $h^{{}^{\prime}}$ is a group homomorphism and the
kernel of $h^{{}^{\prime}}$ is $K_{2\nu+1}$. Hence in order to show that $h$
is a group isomorphism, it suffices to show that for any $\sigma\in
E_{2\nu+1}$, there exist
$k_{1}\in\mathbb{F}_{q}^{*2},~{}k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$,
$\delta_{1}\in\mathbb{F}_{3}^{*}$ and $\pi\in{\rm Aut}(\mathbb{F}_{q})$ such
that

$\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)}.$

Let $\sigma\in E_{2\nu+1}$. And then we will prove that there exist
$k_{1}\in\mathbb{F}_{q}^{*2}$, $k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$,
$\delta_{1}\in\mathbb{F}_{3}^{*}$ and $\pi\in{\rm Aut}(\mathbb{F}_{q^{2}})$
such that $\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\pi)}$.

Let $[(a_{1},a_{2},\ldots,a_{2\nu+1})]\in
V\big{(}Oi\big{(}2\nu+1,q\big{)}\big{)}$ and suppose that
$\sigma([(a_{1},a_{2},\ldots,a_{2\nu+1})])=[(b_{1},b_{2},\ldots,b_{2\nu+1})]$.
Similar to the proof of the case $\delta=0$, we can conclude that

$a_{i}=0\ {\rm if\ and\ only\ if}\ b_{i}=0,\ 1\leq i\leq 2\nu+1,$

and define permutations $\pi_{j}$ of $\mathbb{F}_{q}$ with $\pi_{j}(0)=0$ such
that

$\sigma\big{(}[e_{1}+a_{k}e_{k}]\big{)}=[k_{1}e_{1}+\pi_{k}(a_{k})e_{k}],\ $
$\ \
\sigma\big{(}[e_{1}+a_{\nu+k}f_{k}]\big{)}=[k_{1}e_{1}+\pi_{\nu+k}(a_{\nu+k})f_{k}]\,$

and

$\ \ \ \ \
\sigma\big{(}[e_{1}+a_{2\nu+1}\varepsilon]\big{)}=[k_{1}e_{1}+\pi_{2\nu+1}(a_{2\nu+1})\varepsilon],\
$

where $j=2,\ldots,\nu,\nu+2,\ldots,2\nu+1$, $k=2,\ldots,\nu$ and
$k_{1}^{2}=\pi_{2}(1)^{-1}\pi_{4}(1)^{-1}$.

In the following proof, we need only consider two cases: $\nu=2$ and $\nu>2$.

(1) $\nu=2$. By Theorem 3.2, we know that $\sigma$ carries the vertex
$[(a_{1},a_{2},a_{3},a_{4},a_{5})]$ of $Oi\big{(}5,q\big{)}$ into the vertex

$[(k_{1}\pi(a_{1}),k_{2}\pi(a_{2}),k_{1}^{-1}\pi(a_{3}),k_{2}^{-1}\pi(a_{4}),x_{5})],$

where $k_{1}=\pi_{2}(1)\pi_{4}(1),$ $k_{2}=\pi_{2}(1)$,
$\pi=\pi_{2}(1)^{-1}\pi_{2}=\pi_{4}(1)^{-1}\pi_{4}$ and $x_{5}$ is to be
determined. In order to determine the value of $x_{5},$ we first determine the
images of $[e_{i}+a_{5}e_{5}]$ and $[f_{i}+a_{5}e_{5}]$ for $i=1,2$ under
$\sigma$. Suppose that
$\sigma([e_{i}+a_{5}\varepsilon])=[k_{i}e_{i}+a_{5}^{{}^{\prime}}\varepsilon]$
and
$\sigma([f_{i}+a_{5}\varepsilon])=[k_{i}^{-1}f_{i}+a_{5}^{{}^{\prime}}\varepsilon]$
for $i=1,2$.

Suppose that
$\sigma([(0,0,1,-a_{5},1)])=[(0,0,k_{1}^{-1},k_{2}^{-1}\pi(a_{5}),x)]$ and
$\sigma([(0,0,1,-a_{5},-1)])=[(0,0,\allowbreak
k_{1}^{-1},-k_{2}^{-1}\pi(a_{5}),y)]$. Since
$[(0,0,1,-a_{5},1)]$—$[(1,0,0,0,-1)]$, we have
$[(0,0,k_{1}^{-1},-k_{2}^{-1}\pi(a_{5}),x)]$ — $[(k_{1},0,0,0,\pi_{5}(-1))],$
i.e., $x=-\pi_{5}(-1)^{-1}$. Since $[(0,1,0,0,a_{5})]$ — $[(0,0,1,-a_{5},1)]$,
we have $[(0,k_{2},0,0,a_{5}^{{}^{\prime}})]$ —
$[(0,0,k_{1}^{-1},-k_{2}^{-1}\pi(a_{5}),-\pi_{5}(-1)^{-1})]$, i.e.,
$a_{5}^{{}^{\prime}}=-\pi_{5}(-1)\pi(a_{5})$. Since $[(0,0,1,-a_{5},-1)]$ —
$[(1,0,0,0,1)]$, we have $[(0,0,k_{1}^{-1},-k_{2}^{-1}\pi(a_{5}),y)]$ —
$[(k_{1},0,0,0,\pi_{5}(1))]$, i.e., $y=-\pi_{5}(1)^{-1}$. Since
$[(0,1,0,0,-a_{5})]$ — $[(0,0,1,-a_{5},-1)]$, we have
$[(0,k_{2},0,0,\pi_{5}(-1)\pi(a_{5}))]$ —
$[(0,0,k_{1}^{-1},-k_{2}^{-1}\pi(a_{5}),-\pi_{5}(1)^{-1})]$, i.e.,
$\pi_{5}(1)=-\pi_{5}(-1)$. Thus
$\sigma([(0,1,0,0,a_{5})])=[(0,k_{2},0,0,\pi_{5}(1)\pi(a_{5}))].$

Similarly, we can prove that
$\sigma([(0,0,0,1,a_{5})])=[(0,0,0,k_{2}^{-1},\pi_{5}(1)\pi(a_{5}))]$,
$\sigma([(1,0,0,0,a_{5})])=[(k_{1},0,0,0,\pi_{5}(1)\pi(a_{5}))]$ and
$\sigma([(0,0,1,0,a_{5})])=[(0,0,k_{1}^{-1},0,\pi_{5}(1)\pi(a_{5}))]$.

Since $[(1,0,0,0,1)]$ — $[(0,0,1,0,-1)]$, it follows that
$[(k_{1},0,0,0,\pi_{5}(1))]$ — $[(0,0,k_{1}^{-1},0,\pi_{5}(1))]$, i.e.,
$\pi_{5}(1)^{2}=1$. Hence we have $\pi_{5}(1)\in F_{3}^{*}$. It is easy to
check that

$\sigma([(a_{1},a_{2},a_{3},a_{4},,a_{5})])=[(k_{1}\pi(a_{1}),k_{2}\pi(a_{2}),k_{1}^{-1}\pi(a_{3}),k_{2}^{-1}\pi(a_{4}),\pi_{5}(1)\pi(a_{5}))].$

Lastly, Let
$k_{1}=\pi_{2}(1)\pi_{4}(1),k_{2}=\pi_{2}(1),\delta_{1}=\pi_{5}(1)$ and
$\pi=\pi_{2}(1)^{-1}\pi_{2}=\pi_{4}(1)^{-1}\pi_{4}=\pi_{5}(1)^{-1}\pi_{5}$.
Then for $A=\big{(}a_{ij}\big{)}_{m\times 5}\in
V\big{(}Oi\big{(}5,q\big{)}\big{)}$,

$\sigma\big{(}A\big{)}=\left(\begin{array}[]{ccccc}k_{1}\pi(a_{11})&k_{2}\pi(a_{12})&k_{1}^{-1}\pi(a_{13})&k_{2}^{-1}\pi(a_{14})&\delta_{1}\pi(a_{15})\\\
\vdots&\vdots&\vdots&\vdots&\vdots\\\
k_{1}\pi(a_{m1})&k_{2}\pi(a_{m2})&k_{1}^{-1}\pi(a_{m3})&k_{2}^{-1}\pi(a_{m4})&\delta_{1}\pi(a_{15})\end{array}\right)=\sigma_{(k_{1},k_{2},\delta_{1},\pi)}\big{(}A\big{)}.$

So $\sigma=\sigma_{(k_{1},k_{2},\delta_{1},\pi)}.$

(2) $\nu>2$. Similar to the case of $\nu=2$, we can prove that there exist
$k_{1}\in\mathbb{F}_{q}^{*2},k_{2}\ldots,k_{\nu}\in\mathbb{F}_{q}^{*},\delta_{1}\in\mathbb{F}_{3}^{*2}$
and $\pi\in{\rm Aut}(\mathbb{F}_{q})$ such that
$\sigma=\sigma_{(k_{1},k_{2}\ldots,k_{\nu},\delta_{1},\pi)}.$

######  Corollary 3.5.

Let $\nu\geq 2$ and $\mathbb{F}_{q}$ be a field of characteristic $p$. Then

$\big{|}{\rm
Aut}\big{(}Oi\big{(}2\nu+1,q\big{)}\big{)}\big{|}=\left\\{\begin{array}[]{ll}\frac{1}{2}q^{\nu^{2}}\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}-1)\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}+1)[\mathbb{F}_{q}:\mathbb{F}_{p}],&\textrm{\textmu\textpm}~{}-1\in\mathbb{F}_{q}^{*2},\\\
q^{\nu^{2}}\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}-1)\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}+1)[\mathbb{F}_{q}:\mathbb{F}_{p}],&\textrm{\textmu\textpm}~{}-1\notin\mathbb{F}_{q}^{*2},\end{array}\right.$

###### Proof.

The proof is similar to that of Corollary 3.3.

Finally, we will discuss the case of $\delta=2$. Let $S=S_{2\nu+2,\Delta}$ and
$\Delta={\rm diag}(1,-z)$ with $-1\in\mathbb{F}_{q}^{*2}$. In what follows, we
write $\kappa$ for $e_{2\nu+2}$. Then

$e_{i}{S_{2\nu+2}}{{}^{t}\\!f_{i}}=\varepsilon
S_{2\nu+2}{{}^{t}\\!\varepsilon}=1,\kappa
S_{2\nu+2}{{}^{t}\\!\kappa}=-z,e_{i}{S_{2\nu+2}}{{}^{t}\\!e_{j}}=f_{i}{S_{2\nu+2}}{{}^{t}\\!f_{j}}=0$
for $1\leq i,j\leq\nu$

and

$e_{i}{S_{2\nu+2}}{{}^{t}\\!f_{j}}=e_{i}{S_{2\nu+2}}{{}^{t}\\!\varepsilon}=f_{i}{S_{2\nu+2}}{{}^{t}\\!\varepsilon}=e_{i}{S_{2\nu+2}}{{}^{t}\\!\kappa}=f_{i}{S_{2\nu+2}}{{}^{t}\\!\kappa}=0$
for $i\neq j,1\leq i,j\leq\nu.$

Similar to the case of $\delta=0$, let $\varphi_{2\nu+2}$ be the natural
action of ${\rm Aut}(\mathbb{F}_{q})$ on the group
$\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*}\times\mathbb{F}_{3}^{*}$
defined by
$\varphi_{2\nu+2}(\pi)((k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2}))=(\pi(k_{1}),\pi(k_{2}),\ldots,\pi(k_{\nu}),\pi(\delta_{1}),\pi(\delta_{2}))$
and
$\big{(}\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*}\times\mathbb{F}_{3}^{*}\big{)}\rtimes_{\varphi_{2\nu+2}}{\rm
Aut}\big{(}\mathbb{F}_{q}\big{)}$ be a semidirect product group corresponding
to $\varphi_{2\nu+2}$. Moreover, we define maps
$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)}$ from
$V\big{(}Oi\big{(}2\nu+2,q\big{)}\big{)}$ to itself by

$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)}(A)=\sigma_{\pi}(A)\cdot{\rm
diag}(k_{1},k_{2},\ldots,k_{\nu},k_{1}^{-1},k_{2}^{-1},\ldots,k_{\nu}^{-1},\delta_{1},\delta_{2}),$

where
$k_{1}\in\mathbb{F}_{q}^{*2},k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*},\delta_{1},\delta_{2}\in\mathbb{F}_{3}^{*}$
and $\pi\in{\rm Aut}(\mathbb{F}_{q})$. It is easy to prove that
$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)}\in{\rm
Aut}\big{(}Oi\big{(}2\nu+2,q\big{)}\big{)}$.

######  Theorem 3.6.

Let $\nu\geq 2$ and $E_{2\nu+2}$ be the subgroup of ${\rm
Aut}\big{(}Oi\big{(}2\nu+2,q\big{)}\big{)}$ defined as follows:

$\big{\\{}\sigma\in{\rm
Aut}\big{(}Oi\big{(}2\nu+2,q\big{)}\big{)}:\sigma([e_{i}])=[e_{i}],\sigma([f_{i}])=[f_{i}],\sigma([\varepsilon])=[\varepsilon]\
{\rm and}\ \sigma([\kappa])=[\kappa]\ {\rm for}\ 1\leq i\leq\nu\big{\\}}.$

Then ${\rm
Aut}\big{(}Oi\big{(}2\nu+2,q\big{)}\big{)}=PO_{2\nu+2}\big{(}\mathbb{F}_{q}\big{)}\cdot
E_{2\nu+2}$. Moreover, let

$\displaystyle
h:\left.{\big{(}(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*}\times\mathbb{F}_{3}^{*})\rtimes_{\varphi_{2\nu+2}}{\rm
Aut}(\mathbb{F}_{q})\big{)}}\middle/K_{2\nu+2}\right.$
$\displaystyle\longrightarrow$ $\displaystyle E_{2\nu+2}$
$\displaystyle(k_{1},k_{2},\ldots,k_{\nu},\delta,\pi)K_{2\nu+2}$
$\displaystyle\longrightarrow$
$\displaystyle\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)},$

where $1_{\mathbb{F}_{q}}$ is an identity element of ${\rm
Aut}(\mathbb{F}_{q})$, and
$K_{2\nu+2}=\\{(c,c,\ldots,c,c,c,1_{\mathbb{F}_{q}})\in(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*}\times\mathbb{F}_{3}^{*})\rtimes_{\varphi_{2\nu+2}}{\rm
Aut}(\mathbb{F}_{q}):c=\pm 1\\}$. Then $h$ is an isomorphism of groups.

###### Proof.

Similar to Theorem 3.2, we can show that ${\rm
Aut}\big{(}Oi\big{(}2\nu+2,q\big{)}\big{)}=PO_{2\nu+2}\big{(}\mathbb{F}_{q}\big{)}\cdot
E_{2\nu+2}$. Define a map:

$\displaystyle
h^{{}^{\prime}}:(\mathbb{F}_{q}^{*2}\times\mathbb{F}_{q}^{*}\times\cdots\times\mathbb{F}_{q}^{*}\times\mathbb{F}_{3}^{*}\times\mathbb{F}_{3}^{*})\rtimes_{\varphi_{2\nu+2}}{\rm
Aut}(\mathbb{F}_{q})$ $\displaystyle\longrightarrow$ $\displaystyle
E_{2\nu+2}$
$\displaystyle(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)$
$\displaystyle\longrightarrow$
$\displaystyle\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)}.$

By the definition of
$\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)}$, it is easy
to prove that $h^{{}^{\prime}}$ is a group homomorphism and the kernel of
$h^{{}^{\prime}}$ is $K_{2\nu+2}$. Hence in order to show that $h$ is a group
isomorphism, it suffices to show that for any $\sigma\in E_{2\nu+2}$, there
exist $k_{1}\in\mathbb{F}_{q}^{*2},k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$,
$\delta_{1},\delta_{2}\in\mathbb{F}_{3}^{*}$ and $\pi\in{\rm
Aut}(\mathbb{F}_{q})$ such that

$\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)}.$

Let $\sigma\in E_{2\nu+2}$. And then we will prove that there exist
$k_{1}\in\mathbb{F}_{q}^{*2}$, $k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$,
$\delta_{1},\delta_{2}\in\mathbb{F}_{3}^{*}$ and $\pi\in{\rm
Aut}(\mathbb{F}_{q^{2}})$ such that
$\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)}$.

Let $[(a_{1},a_{2},\ldots,a_{2\nu+1},a_{2\nu+2})]\in
V\big{(}Oi\big{(}2\nu+2,q\big{)}\big{)}$ and suppose that
$\sigma([(a_{1},a_{2},\ldots,a_{2\nu+1},a_{2\nu+2})])=[(b_{1},b_{2},\ldots,b_{2\nu+1},b_{2\nu+2})]$.
Similar to the proof of the case $\delta=0$ and $\delta=1$, we can conclude
that

$a_{i}=0\ {\rm if\ and\ only\ if}\ b_{i}=0\ {\rm for}\ 1\leq i\leq 2\nu+2,$

and define permutations $\pi_{j}$ of $\mathbb{F}_{q}$ with $\pi_{j}(0)=0$ such
that
$\sigma\big{(}[e_{1}+a_{k}e_{k}]\big{)}=[k_{1}e_{1}+\pi_{k}(a_{k})e_{k}],$
$\sigma\big{(}[e_{1}+a_{\nu+k}f_{k}]\big{)}=[k_{1}e_{1}+\pi_{\nu+k}(a_{\nu+k})f_{k}],$
$\sigma\big{(}[e_{1}+a_{2\nu+1}\varepsilon]\big{)}=[k_{1}e_{1}+\pi_{2\nu+1}(a_{2\nu+1})\varepsilon]$
and
$\sigma\big{(}[e_{1}+a_{2\nu+2}\kappa]\big{)}=[k_{1}e_{1}+\pi_{2\nu+2}(a_{2\nu+2})\kappa],$
where $k=2,\ldots,\nu$ and $k_{1}^{2}=\pi_{2}(1)^{-1}\pi_{4}(1)^{-1}$.

In the following proof, we need only consider two cases: $\nu=2$ and $\nu>2$.

(1) $\nu=2$. By Theorem 3.4, we know that $\sigma$ carries the vertex
$[(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6})]$ of $Oi\big{(}6,q\big{)}$ into the
vertex

$[(k_{1}\pi(a_{1}),k_{2}\pi(a_{2}),k_{1}^{-1}\pi(a_{3}),k_{2}^{-1}\pi(a_{4}),\delta_{1}\pi(a_{5}),x_{6})],$

where $k_{1}^{-1}k_{2}=\pi_{2}(1)$,
$k_{1}^{2}=\pi_{2}(1)^{-1}\pi_{4}(1)^{-1}$, $\delta_{1}=\pi_{5}(1)$,
$\pi=\pi_{2}(1)^{-1}\pi_{2}=\pi_{4}(1)^{-1}\pi_{4}=\pi_{5}(1)^{-1}\pi_{5}$ and
$x_{6}$ is to be determined. In order to determine the value of $x_{6},$ we
first determine the images of $[e_{i}+a_{6}\kappa]$, $[f_{i}+a_{6}\kappa]$ and
$[\varepsilon+a_{6}\kappa]$, $i=1,2$ under $\sigma$. Suppose that
$\sigma([e_{i}+a_{6}\kappa])=[k_{i}e_{i}+a_{6}^{{}^{\prime}}\kappa]$,
$\sigma([f_{i}+a_{6}\kappa])=[k_{i}^{-1}f_{i}+a_{6}^{{}^{\prime}}\kappa]$,
$i=1,2$, and
$\sigma([\varepsilon+a_{6}\kappa])=[\delta_{1}\varepsilon+a_{6}^{{}^{\prime}}\kappa]$.

Suppose that
$\sigma([e_{3}+a_{6}e_{4}+z^{-1}\kappa])=[k_{1}e_{3}+k_{2}^{-1}\pi(a_{6})e_{4}+x\kappa]$.
Since $[e_{3}+a_{6}e_{4}+z^{-1}\kappa]$— $[e_{1}+\kappa]$, we have
$[k_{1}^{-1}e_{3}+k_{2}^{-1}\pi(a_{6})e_{4}+x\kappa]$—$[k_{1}e_{1}+\pi_{6}(1)\kappa]$,
i.e., $x=z^{-1}\pi_{6}(1)^{-1}$. Since $[e_{2}+a_{6}\kappa]$ —
$[e_{3}+a_{6}e_{4}+z^{-1}\kappa]$, we have
$[k_{2}e_{2}+a_{6}^{{}^{\prime}}\kappa]$ —
$[k_{1}^{-1}e_{3}+k_{2}^{-1}\pi(a_{6})e_{4}+z^{-1}\pi_{6}(1)^{-1}\kappa]$,
i.e., $a_{6}^{{}^{\prime}}=\pi_{6}(1)\pi(a_{6})$. Thus
$\sigma([e_{2}+a_{6}\kappa])=[k_{2}e_{2}+\pi_{6}(1)\kappa]$.

Similarly, we can prove that
$\sigma([e_{4}+a_{6}\kappa])=[k_{2}^{-1}e_{4}+\pi_{6}(1)\pi(a_{6})\kappa],$
$\sigma([e_{1}+a_{6}\kappa])=[k_{1}e_{1}+\pi_{6}(1)\pi(a_{6})\kappa],$
$\sigma([e_{3}+a_{6}\kappa])=[k_{1}^{-1}e_{3}+\pi_{6}(1)\pi(a_{6})\kappa]$ and
$\sigma([\varepsilon+a_{6}\kappa])=[\delta_{1}\varepsilon+\pi_{6}(1)\pi(a_{6})\kappa].$

We know that $[e_{2}+\kappa]$—$[e_{4}+z^{-1}\kappa]$, and then
$[k_{2}e_{2}+\pi_{6}(1)\kappa]$—$[k_{2}^{-1}e_{4}+\pi(z^{-1})\pi_{6}(1)\kappa]$.
Thus there exists $\delta\in\mathbb{F}_{3}^{*}$ such that
$\pi_{6}(1)=\delta\sqrt{\pi(z)z^{-1}}.$ Moreover, it is easy to check that

$\sigma([(a_{1},a_{2},a_{3},a_{4},,a_{5},a_{6})])=[(k_{1}\pi(a_{1}),k_{2}\pi(a_{2}),k_{1}^{-1}\pi(a_{3}),k_{2}^{-1}\pi(a_{4}),\delta_{1}\pi(a_{5}),\pi_{6}(1)\pi(a_{6}))].$

Lastly, Let $k_{1}^{-1}k_{2}=\pi_{2}(1)$,
$k_{1}=\pi_{2}(1)^{-1}\pi_{4}(1)^{-1}$, $\delta_{1}=\pi_{5}(1)$,
$\delta_{2}=\delta$ and
$\pi=\pi_{2}(1)^{-1}\pi_{2}=\pi_{4}(1)^{-1}\pi_{4}=\pi_{5}(1)^{-1}\pi_{5}=\pi_{6}(1)^{-1}\pi_{6}$.
Then for $A=\big{(}a_{ij}\big{)}_{m\times 6}\in
V\big{(}Oi\big{(}6,q\big{)}\big{)}$,

$\sigma\big{(}A\big{)}=\left(\begin{array}[]{cccccc}k_{1}\pi(a_{11})&k_{2}\pi(a_{12})&k_{1}^{-1}\pi(a_{13})&k_{2}^{-1}\pi(a_{14})&\delta_{1}\pi(a_{15})&\delta_{2}\sqrt{\pi(z)z^{-1}}\pi(a_{16}))\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
k_{1}\pi(a_{m1})&k_{2}\pi(a_{m2})&k_{1}^{-1}\pi(a_{m3})&k_{2}^{-1}\pi(a_{m4})&\delta_{1}\pi(a_{15})&\delta_{2}\sqrt{\pi(z)z^{-1}}\pi(a_{m6}))\end{array}\right)$
$=\,\sigma_{(k_{1},k_{2},\delta_{1},\delta_{2},\pi)}\big{(}A\big{)}.\ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

So $\sigma=\sigma_{(k_{1},k_{2},\delta_{1},\delta_{2},\pi)}.$

(2) $\nu>2$. Similar to the case of $\nu=2$, we can prove that there exist
$k_{1}\in\mathbb{F}_{q}^{*2},k_{2}\ldots,k_{\nu}\in\mathbb{F}_{q}^{*},\delta_{1},\delta_{2}\in\mathbb{F}_{3}^{*2}$
and $\pi\in{\rm Aut}(\mathbb{F}_{q})$ such that
$\sigma=\sigma_{(k_{1},k_{2}\ldots,k_{\nu},\delta_{1},\delta_{2},\pi)}.$

######  Corollary 3.7.

Let $\nu\geq 2$ and $F_{q}$ be a field of characteristic $p$. Then

$\big{|}{\rm
Aut}\big{(}Oi\big{(}2\nu+2,q\big{)}\big{)}\big{|}=\frac{1}{2}q^{\nu(\nu+1)}\mathop{\prod}\limits_{i=1}^{\nu}(q^{i}-1)\mathop{\prod}\limits_{i=1}^{\nu+1}(q^{i}+1)[\mathbb{F}_{q}:\mathbb{F}_{p}].$

###### Proof.

The proof is similar to that of Corollary 3.3.

##  4 The action of Automorphism groups of orthogonal inner graphs of odd
characteristic

######  Theorem 4.1.

Let $Oi\big{(}2\nu+\delta,q\big{)}$ be the orthogonal inner product graph over
$\mathbb{F}_{q}$ and
$\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)\neq\varnothing$ for
$1\leq m<2\nu+\delta$. Then
$\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$ is exactly one orbit of
$V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action of ${\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$.

###### Proof.

By [6, Lemma 6.4] and Proposition 2.4, we know that for any
$A,B\in\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$, there exists
$T\in\mathcal{F}_{2\nu+\delta,\Delta}(q)$ such that
$\sigma_{T}(A)=\sigma_{T}(B).$ Thus ${\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ is transitive on
$\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$.

In order to prove that $\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$
is exactly one orbit of $V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under
the action of ${\rm Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$, we need
only need to show that $\sigma\in{\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$, we have
$\sigma(\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta))=\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta).$
In the following proof, we will consider three cases: $\delta=0,1$ and $2$.

Let $\delta=0$, $\sigma\in{\rm Aut}\big{(}Oi\big{(}2\nu,q\big{)}\big{)}$ and
$A\in\sigma(\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu,\Delta)$.

By Theorem 3.2, we know that there exist
$k_{1}\in\mathbb{F}_{q}^{*2},~{}k_{2},\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$ and
$\pi\in{\rm Aut}(\mathbb{F}_{q})$ such that
$\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}.$ By the knowledge of the
finite field, we know that for non-square element $z\in\mathbb{F}_{q}$, there
exists $a\in\mathbb{F}_{q}^{*2}$ such that $z=a\pi(z)$. If $AS\,{{}^{t}\\!A}$
is cogredient to $M(m_{1},2s_{1}+\gamma_{1},s_{1},\Gamma_{1})$, then it is
easy to check that $\sigma(A)S\,{{}^{t}\\!(\sigma(A))}$ is cogredient to
$M(m_{1},2s_{1}+\gamma_{1},s_{1},\Gamma_{1}).$ So $A$ and $\sigma(A)$ are the
same type orthogonal subspace. Thus we have
$\sigma(\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu,\Delta))=\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu,\Delta).$

Similar to the case of $\delta=0$, we can prove that
$\sigma(\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta))=\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$
when $\delta=1$ and $\delta=2$.

Thus $\mathcal{M}(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$ is exactly one
orbit of $V\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action of
${\rm Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$.

For $A\in Oi\big{(}2\nu+\delta,q\big{)}$, we define
$t(A)=(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$ if the type of $A$ is
$(m,2s+\gamma,s,\Gamma;2\nu+\delta,\Delta)$ and also define

$\phi(A)=\left\\{\begin{array}[]{ll}0,&{\rm\ if\ |A|\ is\ a\
non\text{-square}\ element},\\\ 1,&{\rm\ if\ |A|\ is\ a\ square\
element}.\end{array}\right.$

For $a\in\mathbb{F}_{q}^{*}$, we set $a^{0}=1$ and $a^{1}=a$.

######  Proposition 4.2.

Let $Oi\big{(}2\nu+\delta,q\big{)}$ be the orthogonal inner product graph over
$\mathbb{F}_{q}$ and
$X_{1}\text{---}X_{2},\linebreak[4]Y_{1}\text{---}Y_{2}\in
E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}.$ Then $X_{1}$—$X_{2}$ and
$Y_{1}$—$Y_{2}$ are in the same orbit of
$E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action of
$O_{2\nu+\delta,\Delta}\big{(}\mathbb{F}_{q}\big{)}$ if and only if one of the
following is true$:$ $(1)$ $t(X_{1})=t(Y_{1}),$ $t(X_{2})=t(Y_{2}),$
$t(X_{1}+X_{2})=t(Y_{1}+Y_{2});$ $(2)$ $t(X_{1})=t(Y_{2}),$
$t(X_{2})=t(Y_{1}),$ $t(X_{1}+X_{2})=t(Y_{1}+Y_{2}).$

###### Proof.

($\Rightarrow$) Suppose that $X_{1}$—$X_{2}$ and $Y_{1}$—$Y_{2}$ are in the
same orbit of $E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action
of $O_{2\nu+\delta,\Delta}(\mathbb{F}_{q})$. Then there exists $T\in
O_{2\nu+\delta,\Delta}(\mathbb{F}_{q})$ such that one of the following is
true: $(1)X_{1}T=Y_{1},$ $X_{2}T=Y_{2}$; $(2)X_{1}T=Y_{2},$ $X_{2}T=Y_{1}.$
Without loss of generality we can assume that $X_{1}T=Y_{1}$ and
$X_{2}T=Y_{2}$. Then $t(X_{1})=t(Y_{1})$ and $t(X_{2})=t(Y_{2})$ by
Proposition 2.4 and Theorem 4.1. It is sufficient to prove
$t(X_{1}+X_{2})=t(Y_{1}+Y_{2})$. Let
$t(X_{1})=t(Y_{1})=(m_{1},2s_{1}+\gamma_{1},s_{1},\Gamma_{1};2\nu+\delta,\Delta)$
and
$t(X_{2})=t(Y_{2})=(m_{2},2s_{2}+\gamma_{2},s_{2},\Gamma_{2};2\nu+\delta,\Delta)$.
Then we have

$X_{1}T=\begin{bmatrix}X_{11}\\\ \vdots\\\
X_{1m_{1}}\end{bmatrix}T=\begin{bmatrix}Y_{11}\\\ \vdots\\\
Y_{1m_{1}}\end{bmatrix}=Y_{1}{\rm\ and\ }X_{2}T=\begin{bmatrix}X_{21}\\\
\vdots\\\ X_{2m_{2}}\end{bmatrix}T=\begin{bmatrix}Y_{21}\\\ \vdots\\\
Y_{2m_{2}}\end{bmatrix}=Y_{2}.$

It is easy to check that $(X_{1}+X_{2})T=(Y_{1}+Y_{2})$. By Lemma [6, Lemma
6.4], we have $t(X_{1}+X_{2})=t(Y_{1}+Y_{2})$.

When $X_{1}T=Y_{2}$ and $X_{2}T=Y_{1},$ we conclude similarly that
$t(X_{1})=t(Y_{2}),$ $t(X_{2})=t(Y_{1})$ and $t(X_{1}+X_{2})=t(Y_{1}+Y_{2}).$

($\Leftarrow$) Suppose that one of the following is true: $(1)$
$t(X_{1})=t(Y_{1}),$ $t(X_{2})=t(Y_{2}),$ $t(X_{1}+X_{2})=t(Y_{1}+Y_{2})$;
$(2)$ $t(X_{1})=t(Y_{2}),$ $t(X_{2})=t(Y_{1}),$
$t(X_{1}+X_{2})=t(Y_{1}+Y_{2})$. Without loss of generality we can assume that
$t(X_{1})=t(Y_{1})=(m_{1},r_{1}),$ $t(X_{2})=t(Y_{2})=(m_{2},r_{2})$ and
$t(X_{1}+X_{2})=t(Y_{1}+Y_{2})=(m,r).$ Then there exist matrix representations

$\begin{bmatrix}\alpha_{1}\\\ \vdots\\\ \alpha_{m-m_{2}}\\\ \gamma_{1}\\\
\vdots\\\ \gamma_{m_{1}+m_{2}-m}\end{bmatrix},~{}\begin{bmatrix}\beta_{1}\\\
\vdots\\\ \beta_{m-m_{1}}\\\ \gamma_{1}\\\ \vdots\\\
\gamma_{m_{1}+m_{2}-m}\end{bmatrix}~{}\text{and}~{}\begin{bmatrix}\gamma_{1}\\\
\vdots\\\ \gamma_{m_{1}+m_{2}-m}\end{bmatrix}$

of $X_{1},$ $X_{2}$ and $X_{1}\bigcap X_{2}$ respectively such that

$X_{1}+X_{2}=\begin{bmatrix}\alpha_{1}\\\ \vdots\\\ \alpha_{m-m_{2}}\\\
\beta_{1}\\\ \vdots\\\ \beta_{m-m_{1}}\\\ \gamma_{1}\\\ \vdots\\\
\gamma_{m_{1}+m_{2}-m}\end{bmatrix},$

and

$\alpha_{i}S\,{{}^{t}\\!\beta_{j}}=0,\>\alpha_{i}S\,{{}^{t}\\!\gamma_{k}}=0,\>\beta_{j}S\,{{}^{t}\\!\gamma_{k}}=0$

where $1\leq i\leq m-m_{2}$, $1\leq j\leq m-m_{1}$ and $1\leq k\leq
m_{1}+m_{2}-m$. It is easy to verify that the type of $t(X_{1}\bigcap X_{2})$
is determine by $X_{1}$ and $X_{2}$. We can choose a suitable basis

$\alpha_{1},\ldots,\alpha_{m-m_{2}},~{}\beta_{1},\ldots,\beta_{m-m_{1}},~{}\gamma_{1},\ldots,\linebreak[4]\gamma_{m_{1}+m_{2}-m}$

of $X_{1}+X_{2}$ such that $(X_{1}+X_{2})S\,{{}^{t}(X_{1}+X_{2})}=$

${\rm
diag}(x_{1},\ldots,x_{r_{1}-1},z^{\phi(X_{1})},0,\ldots,0,x_{1},\ldots,x_{r_{2}-1},z^{\phi(X_{2})},0,\ldots,0,x_{1},\ldots,x_{r_{3}-1},z^{\phi(X_{3})}0,\ldots,0),$

where $x_{h}=1$, $1\leq h<\max\\{r_{1},\>r_{2},\>r_{3}\\}$.

Similarly, we can prove that there exists a matrix representation

$Y_{1}+Y_{2}=\begin{bmatrix}\alpha_{1}^{{}^{\prime}}\\\ \vdots\\\
\alpha_{m-m_{2}}^{{}^{\prime}}\\\ \beta_{1}^{{}^{\prime}}\\\ \vdots\\\
\beta_{m-m_{1}}^{{}^{\prime}}\\\ \gamma_{1}^{{}^{\prime}}\\\ \vdots\\\
\gamma_{m_{1}+m_{2}-m}^{{}^{\prime}}\end{bmatrix},$

of $Y_{1}+Y_{2}$ such that

$Y_{1}=\begin{bmatrix}\alpha_{1}^{{}^{\prime}}\\\ \vdots\\\
\alpha_{m-m_{2}}^{{}^{\prime}}\\\ \gamma_{1}^{{}^{\prime}}\\\ \vdots\\\
\gamma_{m_{1}+m_{2}-m}^{{}^{\prime}}\end{bmatrix},~{}~{}Y_{2}=\begin{bmatrix}\beta_{1}^{{}^{\prime}}\\\
\vdots\\\ \beta_{m-m_{1}}^{{}^{\prime}}\\\ \gamma_{1}^{{}^{\prime}}\\\
\vdots\\\ \gamma_{m_{1}+m_{2}-m}^{{}^{\prime}}\end{bmatrix},$

and $(Y_{1}+Y_{2})S\,{{}^{t}(Y_{1}+Y_{2})}=$

${\rm
diag}(y_{1},\ldots,y_{r_{1}-1},z^{\phi(Y_{1})},0,\ldots,0,y_{1},\ldots,y_{r_{2}-1},z^{\phi(Y_{2})},0,\ldots,0,y_{1},\ldots,y_{r_{3}-1},z^{\phi(Y_{3})}0,\ldots,0),$

where $y_{h}=1$, $1\leq h<\max\\{r_{1},\>r_{2},\>r_{3}\\}$ and
$\phi(X_{i})=\phi(Y_{i}),i=1,2,3$.

By [6, Lemma 6.8], there exists $T\in
O_{2\nu+\delta,\Delta}\big{(}\mathbb{F}_{q}\big{)}$ such that
$(X_{1}+X_{2})T=Y_{1}+Y_{2}$, $X_{1}T=Y_{1}$ and $X_{2}T=Y_{2}$. So
$X_{1}$—$X_{2}$ and $Y_{1}$—$Y_{2}$ are in the same orbit of
$E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action of
$O_{2\nu+\delta,\Delta}\big{(}\mathbb{F}_{q}\big{)}$.

When $t(X_{1})=t(Y_{2}),$ $t(X_{2})=t(Y_{1})$ and
$t(X_{1}+X_{2})=t(Y_{1}+Y_{2}),$ we conclude similarly that $X_{1}$—$X_{2}$
and $Y_{1}$—$Y_{2}$ are in the same orbit of
$E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action of
$O_{2\nu+\delta,\Delta}\big{(}\mathbb{F}_{q}\big{)}$.

######  Theorem 4.3.

Let $Oi\big{(}2\nu+\delta,q\big{)}$ be the orthogonal inner product graph over
$\mathbb{F}_{q}$ and $X_{1}\text{---}X_{2},Y_{1}\text{---}Y_{2}\in
E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}.$ Then $X_{1}$—$X_{2}$ and
$Y_{1}$—$Y_{2}$ are in the same orbit of
$E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action of ${\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ if and only if one of the
following is true$:$ $(1)$ $t(X_{1})=t(Y_{1}),$ $t(X_{2})=t(Y_{2}),$
$t(X_{1}+X_{2})=t(Y_{1}+Y_{2});$ $(2)$ $t(X_{1})=t(Y_{2}),$
$t(X_{2})=t(Y_{1}),$ $t(X_{1}+X_{2})=t(Y_{1}+Y_{2}).$

###### Proof.

($\Rightarrow$) Suppose that $X_{1}$—$X_{2}$ and $Y_{1}$—$Y_{2}$ are in the
same orbit of $E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action
of ${\rm Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}.$ Then there exists
$\sigma\in{\rm Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ such that one
of the following is true: $\sigma(X_{1})=Y_{1},$ $\sigma(X_{2})=Y_{2}$;
$\sigma(X_{1})=Y_{2},$ $\sigma(X_{2})=Y_{1}.$ Without loss of generality we
can assume that $\sigma(X_{1})=Y_{1}$ and $\sigma(X_{2})=Y_{2}$. Then
$t(X_{1})=t(Y_{1})$ and $t(X_{2})=t(Y_{2})$ by Theorem 4.1. It is sufficient
to prove $t(X_{1}+X_{2})=t(Y_{1}+Y_{2})$. In order to prove
$t(X_{1}+X_{2})=t(Y_{1}+Y_{2}),$ we need only consider three cases:
$\delta=0,1$ and 2. We first consider the case $\delta=0$. Let $\delta=0$. By
Theorem 3.2, we know that there exist
$k_{1}\in\mathbb{F}_{q}^{*2},k_{2}\ldots,k_{\nu}\in\mathbb{F}_{q}^{*}$ and
$\pi\in{\rm Aut}(\mathbb{F}_{q})$ such that
$\sigma=\sigma_{(k_{1},k_{2},\ldots,k_{\nu},\pi)}.$ By Proposition 4.2, we
know that $X_{1}+X_{2}$ and $\sigma_{\pi^{-1}}\sigma(X_{1}+X_{2})$ are the
same type orthogonal subspace. It is easy to check that
$M(m,2s+\gamma,s,\Gamma)$ is cogredient to
$\sigma_{\pi}(M(m,2s+\gamma,s,\Gamma))$. So
$\sigma_{\pi^{-1}}\sigma(X_{1}+X_{2})$ and $\sigma(X_{1}+X_{2})$ are the same
type orthogonal subspace. Thus $X_{1}+X_{2}$ and
$\sigma(X_{1}+X_{2})=Y_{1}+Y_{2}$ are the same type orthogonal subspace.

Similar to the case of $\delta=0$, we can prove that $X_{1}+X_{2}$ and
$Y_{1}+Y_{2}$ are the same type orthogonal subspace when $\delta=1$ and
$\delta=2$.

($\Leftarrow$) Suppose that one of the following is true: $(1)$
$t(X_{1})=t(Y_{1}),$ $t(X_{2})=t(Y_{2}),$ $t(X_{1}+X_{2})=t(Y_{1}+Y_{2})$;
$(2)$ $t(X_{1})=t(Y_{2}),$ $t(X_{2})=t(Y_{1}),$
$t(X_{1}+X_{2})=t(Y_{1}+Y_{2}).$

By Proposition 2.4 and 4.1, it is easy to check that there exist $T\in
O_{2\nu+\delta,\Delta}(\mathbb{F}_{q})$ and $\sigma_{T}\in{\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ such that

$\sigma_{T}(X_{1})=X_{1}T=Y_{1}~{}~{}{\rm
and}~{}~{}\sigma_{T}(X_{2})=X_{2}T=Y_{2}$

or

$\sigma_{T}(X_{1})=X_{1}T=Y_{2}~{}~{}{\rm
and}~{}~{}\sigma_{T}(X_{2})=X_{2}T=Y_{1}.$

Thus $X_{1}$—$X_{2}$ and $Y_{1}$—$Y_{2}$ are in the same orbit of
$E\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$ under the action of ${\rm
Aut}\big{(}Oi\big{(}2\nu+\delta,q\big{)}\big{)}$.

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