By convexity, we can restrict the set of $\lambda$ that we must optimize over
to the extremal values, which gives:

$\displaystyle\Phi_{\rho}(\bm{x})$
$\displaystyle=\begin{cases}\frac{1}{2}(1+\bar{\bm{x}}\cdot\bm{r})&\text{if
}\bar{\bm{x}}\cdot\bm{r}\geq 0,\\\
\frac{1}{2}(1-\frac{1}{3}\bar{\bm{x}}\cdot\bm{r})&\mbox{
otherwise}.\end{cases}$ (184)

In words, if the Bloch vectors of $\eta^{T}$ and $\rho$ are both located in
the hemisphere agove the plane perpendicular to the Bloch vector of
$\eta^{T}$, then the identity channel ${\cal E}_{1}=\mathcal{I}$ optimizes the
objective function. Otherwise, the channel ${\cal E}_{-\frac{1}{3}}$ which
achieves the best inversion of $\rho$ that can be done covariantly, is
optimal.

##  Appendix H Depolarization conditions

###  H.1 Depolarization conditions via the modes of asymmetry

######  Lemma 27.

Let $\rho_{A}$, $\sigma_{B}$, and $\eta_{R}$ be quantum states on systems $A$,
$B$, and $R$, respectively, where $d_{R}=d_{B}$. We have
$\rho\xrightarrow{G}\sigma$ if for all quantum states $\eta$ on ${\cal H}_{R}$

$\displaystyle\langle\eta,{\cal
E}^{\eta}(\rho)\rangle\geq\langle\eta,\sigma\rangle,$ (185)

for any family of covariant channels ${\cal E}^{\eta}:{\cal S}({\cal
H}_{A})\rightarrow{\cal S}({\cal H}_{B})$ parameterised by $\eta$, where
$\langle A,B\rangle\coloneqq\mathrm{tr}[A^{\dagger}B]$.

###### Proof.

Lemma 3 of the supplemental material of [58] tells us that
$\rho\xrightarrow{G}\sigma$ if and only if

$\displaystyle\Phi_{\eta}(\rho)\geq\mathrm{tr}[\eta^{T}\sigma],\forall\eta.$
(186)

From the dual expression for $\Phi_{\eta}(\rho)$ in Lemma 12, we have the
following lower bound on $\Phi_{\eta}(\rho)$ for any choice of covariant
channel ${\cal E}^{\eta}$:

$\Phi_{\eta}(\rho)\geq\mathrm{tr}[\eta^{T}{\cal E}^{\eta}(\rho)].$ (187)

Eqs. (186) and (187) allow us to generate the following sufficient condition,
such that if we can find any faimily of covariant maps ${\cal E}^{\eta}$ such
that

$\displaystyle\mathrm{tr}[\eta^{T}{\cal
E}^{\eta}(\rho)]\geq\mathrm{tr}[\eta^{T}\sigma],\forall\eta,$ (188)

or, equivalently

$\displaystyle\mathrm{tr}[\eta{\cal
E}^{\eta}(\rho)]\geq\mathrm{tr}[\eta\sigma],\forall\eta,$ (189)

then we are guaranteed $\rho\xrightarrow{G}\sigma$, which completes the proof.
∎

Lemmata 27 and 16 together produce the following corollary:

######  Corollary 28.

We have that $\rho\xrightarrow{G}\sigma$ if for all $\eta\in{\cal S}({\cal
H}_{R})$ ,

$\displaystyle\sum_{\lambda,j}\langle\eta^{\lambda}_{j},{\cal
E}^{\eta}(\rho^{\lambda}_{j})-\sigma_{j}^{\lambda}\rangle\geq 0.$ (190)

for some family of covariant channels ${\cal E}^{\eta}$.

###  H.2 PGM measure-and-prepare channel

For a general group $G$ we have the Pretty Good Measurement [80] POVM:

$\displaystyle M_{\mathrm{pgm}}(g)\coloneqq{\cal
G}(\rho)^{-\frac{1}{2}}\rho(g){\cal G}(\rho)^{-\frac{1}{2}},$ (191)

where $\rho(g)\coloneqq{\cal U}_{g}(\rho)$.

Recall $\sigma^{\lambda}_{j}={\cal E}(\rho^{\lambda}_{j})$ [60] and
$\mathrm{tr}[AB]=\sum_{\lambda,j}\mathrm{tr}[A^{\lambda}_{j}B^{\lambda^{*}}_{j}]$,
we can evaluate the trace product for the PGM measure-and-prepare channel,
which prepares ${\cal U}_{g}(\tau)$ for an arbitrary state $\tau$ when the
$g$th outcome is obtained:

${\cal E}_{\rm pgm}(\rho):=\int dg\ \tau(g)\mathrm{tr}[M_{\rm pgm}(g)\rho],$
(192)

which, from its form, is manifestly covariant.

We then have that

$\displaystyle\mathrm{tr}[\eta{\cal
E}_{\mathrm{pgm}}(\rho)]=\sum_{\lambda,j}\mathrm{tr}[\eta_{j}^{\lambda^{*}}{\cal
E}_{\mathrm{pgm}}(\rho_{j}^{\lambda})]=\sum_{\lambda,j}\int
dg\mathrm{tr}[M_{\mathrm{pgm}}(g)\rho_{j}^{\lambda}]\mathrm{tr}[\eta^{\lambda^{*}}_{j}\tau(g)].$
(193)

We can write each term in the above equation more compactly in terms of the
Hilbert-Schmidt inner product:

$\displaystyle\langle\eta^{\lambda}_{j},{\cal
E}_{\mathrm{pgm}}(\rho_{j}^{\lambda})\rangle=\int
dg\,\langle\eta_{j}^{\lambda},\tau(g)\rangle\langle\rho^{\lambda^{*}}_{j},\overline{\rho}(g)\rangle,$
(194)

where we have defined the scaled variant of $\rho$, $\overline{\rho}$ as

$\displaystyle\overline{\rho}\coloneqq{\cal G}(\rho)^{-\frac{1}{2}}\rho{\cal
G}(\rho)^{-\frac{1}{2}}.$ (195)

We now have the following lemma:

######  Lemma 29.

For any irrep $\lambda$ and irrep component $j$ we have

$\langle\eta^{\lambda}_{j},{\cal
E}_{\mathrm{pgm}}(\rho_{j}^{\lambda})\rangle=f^{\lambda}_{j}(\rho)\langle\eta_{j}^{\lambda},\tau_{j}^{\lambda}\rangle,$
(196)

where we have introduced the functions
$f^{\lambda}_{j}(\rho)\coloneqq\langle\rho^{\lambda^{*}}_{j},\overline{\rho}_{j}^{\lambda^{*}}\rangle$.

###### Proof.

We have:

$\displaystyle\langle\eta^{\lambda}_{j},{\cal
E}_{\mathrm{pgm}}(\rho_{j}^{\lambda})\rangle$ $\displaystyle=\int
dg\,\langle\eta_{j}^{\lambda},\tau(g)\rangle\langle\rho^{\lambda^{*}}_{j},\overline{\rho}(g)\rangle$
(197) $\displaystyle=\int
dg\,\langle\eta^{\lambda}_{j}\otimes\rho^{\lambda^{*}}_{j},\tau(g)\otimes\overline{\rho}(g)\rangle$
(198)
$\displaystyle=\langle\eta^{\lambda}_{j}\otimes\rho^{\lambda^{*}}_{j},\int
dg\,{\cal U}_{g}(\tau\otimes\overline{\rho})\rangle$ (199)
$\displaystyle=\langle\eta^{\lambda}_{j}\otimes\rho^{\lambda^{*}}_{j},{\cal
G}(\tau\otimes\overline{\rho})\rangle$ (200)
$\displaystyle=\sum_{i,\mu}\langle\eta^{\lambda}_{j}\otimes\rho^{\lambda^{*}}_{j},\tau_{i}^{\mu}\otimes\overline{\rho}_{i}^{\mu^{*}}\rangle$
(201)
$\displaystyle=\langle\eta^{\lambda}_{j}\otimes\rho^{\lambda^{*}}_{j},\tau_{j}^{\lambda}\otimes\overline{\rho}_{j}^{\lambda^{*}}\rangle$
(202)
$\displaystyle=\langle\eta^{\lambda}_{j},{\tau}_{j}^{\lambda}\rangle\langle\rho^{\lambda^{*}}_{j},\overline{\rho}_{j}^{\lambda^{*}}\rangle$
(203)
$\displaystyle=\langle\eta^{\lambda}_{j},f^{\lambda}_{j}(\rho){\tau}_{j}^{\lambda}\rangle,$
(204)

where in going from line 200 to line 201 we have made use of lemma 17. This
completes the proof. ∎

We now note the following properties of the functions $f^{\lambda}_{j}(\rho)$,
which hold for any input state $\rho$ and for all modes labelled by
$(\lambda,j)$:

  1. (F1)

(Real-valued).
$f^{\lambda}_{j}(\rho)=f^{\lambda^{*}}_{j}(\rho)=[f^{\lambda}_{j}(\rho)]^{*}$.

  2. (F2)

(Non-negative). $f^{\lambda}_{j}(\rho)\geq 0$.

  3. (F3)

(Trivial irrep). $f^{0}_{0}(\rho)=1$.

###### Proof of (F1).

We note that

$\displaystyle f_{j}^{\lambda}(\rho)=\mathrm{tr}[\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{2}}\rho^{\lambda^{*}}_{j}{\cal G}(\rho)^{-\frac{1}{2}}]$
$\displaystyle=\mathrm{tr}[\rho^{\lambda^{*}}_{j}{\cal
G}(\rho)^{-\frac{1}{2}}\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{2}}]=f_{j}^{\lambda^{*}}(\rho)$ (205)
$\displaystyle=\mathrm{tr}[(\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{2}}\rho^{\lambda^{*}}_{j}{\cal
G}(\rho)^{-\frac{1}{2}})^{\dagger}]=\mathrm{tr}[\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{2}}\rho^{\lambda^{*}}_{j}{\cal
G}(\rho)^{-\frac{1}{2}}]^{*}=[f^{\lambda}_{j}(\rho)]^{*}$ (206)

∎

###### Proof of (F2).

We begin by noting that

$\displaystyle\rho^{\lambda}_{j}\mathcal{G}(\rho)^{-\frac{1}{2}}\rho^{\lambda^{*}}_{j}=\left(\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{4}}\right)\left(\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{4}}\right)^{\dagger}\eqqcolon AA^{\dagger},$ (207)

where we have defined $A\coloneqq\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{4}}$. Regardless of what $A$ is, any operator of the form
$AA^{\dagger}$ is positive. Since $\mathcal{G}(\rho)^{-\frac{1}{2}}$ is a
positive operator as well, this means
$\rho^{\lambda}_{j}\mathcal{G}(\rho)^{-\frac{1}{2}}\rho^{\lambda^{*}}_{j}\mathcal{G}(\rho)^{-\frac{1}{2}}$
must be positive, so we are guaranteed $f^{\lambda}_{j}(\rho)\geq
0,\forall\lambda,j,\rho$.

∎

###### Proof of (F3).

We first note

$\displaystyle\mathcal{G}(\rho)$ $\displaystyle=\sum_{\lambda,j}\int dg\
\mathcal{U}_{g}\left(\sum_{\alpha}\mathrm{tr}({X_{j}^{(\lambda,\alpha)}}^{\dagger}\rho)X_{j}^{(\lambda,\alpha)}\right)$
(208) $\displaystyle=\sum_{\lambda,j,j^{\prime}}\int
dgv^{(\lambda)}_{j^{\prime}j}(g)\sum_{\alpha}\mathrm{tr}({X_{j}^{(\lambda,\alpha)}}^{\dagger}\rho)X_{j^{\prime}}^{(\lambda,\alpha)}$
(209) $\displaystyle=\sum_{\lambda,j,j^{\prime}}\left[\int
dgv^{(\lambda)}_{j^{\prime}j}(g)v^{0}_{0,0}(g)\right]\sum_{\alpha}\mathrm{tr}({X_{j}^{(\lambda,\alpha)}}^{\dagger}\rho)X_{j^{\prime}}^{(\lambda,\alpha)}$
(210)
$\displaystyle=\sum_{\lambda,j,j^{\prime}}\delta_{\lambda,0}\delta_{j^{\prime},0,}\delta_{j,0}\sum_{\alpha}\mathrm{tr}({X_{j}^{(\lambda,\alpha)}}^{\dagger}\rho)X_{j^{\prime}}^{(\lambda,\alpha)}$
(211)
$\displaystyle=\sum_{\alpha}\mathrm{tr}({X_{0}^{(0,\alpha)}}^{\dagger}\rho)X_{0}^{(0,\alpha)}=\rho^{0}_{0}.$
(212)

where in going from line 210 to line 211 we have made use of Equation 107. It
then follows that:

$\displaystyle
f_{0}^{0}(\rho)=\mathrm{tr}[\mathcal{G}(\rho)\mathcal{G}(\rho)^{-\frac{1}{2}}\mathcal{G}(\rho)\mathcal{G}(\rho)^{-\frac{1}{2}}]=\mathrm{tr}[\mathcal{G}(\rho)]=1,$
(213)

which completes the proof.

∎

###  H.3 Derivation of general conditions

######  Lemma 30.

Let us define components $t^{(\lambda,\alpha)}_{j}\coloneqq\langle
X^{(\lambda,\alpha)}_{j},\tau\rangle$ and
$s^{(\lambda,\alpha)}_{j}\coloneqq\langle
X^{(\lambda,\alpha)}_{j},\sigma\rangle$. The state $\rho$ can be transformed
into $\sigma$ covariantly with respect to a symmetry group $G$ (i.e.
$\rho\xrightarrow{G}\sigma$) if there exists a family of valid quantum states

$\displaystyle\left\\{\forall\theta^{(\lambda,\alpha)}_{j}\in[0,2\pi):\tau_{\Theta}=\sum_{\lambda,\alpha,j}t^{(\lambda,\alpha)}_{j}X^{(\lambda,\alpha)}_{j}\right\\},$
(214)

where each $(\lambda,\alpha,j)$-component of $\tau_{\Theta}$ satisfies either:

$\displaystyle f^{\lambda}_{j}(\rho)t^{(\lambda,\alpha)}_{j}$
$\displaystyle=s^{(\lambda,\alpha)}_{j},\text{ or,}$ (215) $\displaystyle
f^{\lambda}_{j}(\rho)\absolutevalue{t^{(\lambda,\alpha)}_{j}}-\absolutevalue{s^{(\lambda,\alpha)}_{j}}$
$\displaystyle\geq 0,\quad
t^{(\lambda,\alpha)}_{j}=e^{i\theta^{(\lambda,\alpha)}_{j}}\absolutevalue{t^{(\lambda,\alpha)}_{j}}.$
(216)

###### Proof.

We begin with Corollary 28 and make the explicit choice for each member in our
family of covariant channels $\\{{\cal E}^{\eta}\\}$ to be a PGM-and-prepare
channel ${\cal E}^{\eta}\coloneqq{\cal E}^{\eta}_{\mathrm{pgm}}$, as defined
in Eq. (192). Combined with Lemma 29 this gives rise to the following
sufficient condition on the transition $\rho\xrightarrow{G}\sigma$:

$\displaystyle\sum_{\lambda,j}\left\\{\langle\eta^{\lambda}_{j},f_{j}^{\lambda}(\rho)\tau_{j}^{\lambda}-\sigma_{j}^{\lambda}\rangle+\langle\eta^{\lambda^{*}}_{j},f_{j}^{\lambda}(\rho)\tau_{j}^{\lambda^{*}}-\sigma_{j}^{\lambda^{*}}\rangle\right\\}\geq
0,\quad\forall\eta.$ (217)

We emphasise that the choice of preparation state $\tau$ can vary with $\eta$,
though this dependence has been suppressed in our notation for clarity. We now
introduce the simplifying notation $\bm{\mu}\coloneqq\\{\lambda,\alpha,j\\}$
and define the following coefficients of the states $\eta,\tau,$ and $\sigma$
in the ITO basis $\\{X^{\bm{\mu}}\\}$:

$\displaystyle n^{\bm{\mu}}$ $\displaystyle\coloneqq\langle
X^{\bm{\mu}},\eta\rangle=\absolutevalue{n^{\bm{\mu}}}e^{i\varphi_{\eta}^{\bm{\mu}}},$
(218) $\displaystyle t^{\bm{\mu}}$ $\displaystyle\coloneqq\langle
X^{\bm{\mu}},\tau\rangle=\absolutevalue{t^{\bm{\mu}}}e^{i\varphi_{\tau}^{\bm{\mu}}},$
(219) $\displaystyle s^{\bm{\mu}}$ $\displaystyle\coloneqq\langle
X^{\bm{\mu}},\sigma\rangle=\absolutevalue{s^{\bm{\mu}}}e^{i\varphi_{\sigma}^{\bm{\mu}}},$
(220)

such that $\eta_{j}^{\lambda}=\sum_{\alpha}n^{\bm{\mu}}X^{\bm{\mu}}$ etc. By
the hermiticity of $\eta$, we have $n^{\bm{\mu}^{*}}\coloneqq
n^{(\lambda^{*},\alpha)}_{j}=\left(n^{(\lambda,\alpha)}_{j}\right)^{*}=(n^{\bm{\mu}})^{*}$
etc. We further introduce the notation

$\displaystyle f^{\bm{\mu}}(\rho)\coloneqq
f^{\lambda}_{j}(\rho),\forall\alpha.$ (221)

Substituting these definitions into Eq. (217) and using the orthonormality of
the ITO basis, $\langle
X^{\bm{\mu}},X^{\bm{\nu}}\rangle=\delta_{\bm{\mu},\bm{\nu}}$, gives the
following sufficient condition:

$\displaystyle\sum_{\bm{\mu}}\left[n^{\bm{\mu}^{*}}\left(f^{\bm{\mu}}(\rho)t^{\bm{\mu}}-s^{\bm{\mu}}\right)+n^{\bm{\mu}}\left(f^{\bm{\mu}}(\rho)t^{\bm{\mu}^{*}}-s^{\bm{\mu}^{*}}\right)\right]\geq
0,\quad\forall\eta.$ (222)

This can instead be written as

$\displaystyle\sum_{\bm{\mu}}\real\\{n^{\bm{\mu^{*}}}\left(f^{\bm{\mu}}(\rho)t^{\bm{\mu}}-s^{\bm{\mu}}\right)\\}=\sum_{\bm{\mu}}\absolutevalue{n^{\bm{\mu^{*}}}}\real\left\\{e^{-i\varphi_{\eta}^{\bm{\mu}}}\left[f^{\bm{\mu}}(\rho)t^{\bm{\mu}}-s^{\bm{\mu}}\right]\right\\}\geq
0,\quad\forall\eta.$ (223)

We can always choose $X^{\bm{0}}\coloneqq\frac{\mathbbm{1}}{\sqrt{d}}$, which
means all other ITO basis elements must be traceless. Since we must choose
$\tau$ to be a valid quantum state, this means we must assign

$\displaystyle\mathrm{tr}(\tau)=\sum_{\bm{\mu}}t^{\bm{\mu}}\mathrm{tr}(X^{\bm{\mu}})=\sqrt{d}t^{\bm{0}}=1.$
(224)

As $\sigma$ is a valid quantum state, by similar logic we conclude that
$\sqrt{d}s^{\bm{0}}=1$. Due to (F3), this means Eq. (223) reduces to

$\displaystyle\sum_{\bm{\mu}\neq\bm{0}}\absolutevalue{n^{\bm{\mu^{*}}}}\real\left\\{e^{-i\varphi_{\eta}^{\bm{\mu}}}\left[f^{\bm{\mu}}(\rho)t^{\bm{\mu}}-s^{\bm{\mu}}\right]\right\\}\geq
0,\quad\forall\eta.$ (225)

Given any $\bm{\nu}\neq\bm{0}$, there exists a valid reference state
$\eta=\sum_{\bm{\mu}}n^{\bm{\mu}}X^{\bm{\mu}}$ where

$\displaystyle n^{\bm{\mu}}=\begin{cases}\frac{1}{\sqrt{d}}&\text{ for
}\bm{\mu}=\bm{0}\\\ n\text{ such that }\frac{1}{2d}\geq n>0&\text{ for
}\bm{\mu}=\bm{\nu},\bm{\nu^{*}}\\\ 0&\text{ otherwise,}\end{cases}$ (226)

since these assignments evidently lead to $\eta$ being Hermitian and trace 1,
and we can further verify, for any pure state $\ket{\psi}$ of the reference
system, that

$\displaystyle\bra{\psi}\eta\ket{\psi}=\sum_{\bm{\mu}}n^{\bm{\mu}}\bra{\psi}X^{\bm{\mu}}\ket{\psi}=\frac{1}{d}-n(\bra{\psi}X^{\bm{\nu}}\ket{\psi}+\bra{\psi}X^{\bm{\nu^{*}}}\ket{\psi})\geq\frac{1}{d}-2n\geq\frac{1}{d}-\frac{2}{2d}=0\Rightarrow\eta\geq
0.$ (227)

The only way that Eq. (225) can be satisfied for reference states with
components assigned according to Eq. (226) is if for each
$\bm{\mu}$-component, we have

$\displaystyle\real\left\\{e^{-i\varphi_{\eta}^{\bm{\mu}}}\left[f^{\bm{\mu}}(\rho)t^{\bm{\mu}}-s^{\bm{\mu}}\right]\right\\}\geq
0,\quad\forall\eta.$ (228)

Recall that we are free to choose $\tau$ as we like for every $\eta$, and our
aim here is to derive a set of conditions that are independent of $\eta$. For
a given component $\bm{\mu}$, one possible way of satisfying Eq. (228)
independently of $\eta$ is if we can choose a valid quantum state
$\tau=\sum_{\bm{\mu}}t^{\bm{\mu}}X^{\bm{\mu}}$ such that

$\displaystyle f^{\bm{\mu}}(\rho)t^{\bm{\mu}}=s^{\bm{\mu}}.$ (229)

Alternatively, note that we can rewrite Eq. (228) as follows:

$\displaystyle
f^{\bm{\mu}}(\rho)\absolutevalue{t^{\bm{\mu}}}\cos{(\varphi_{\tau}^{\bm{\mu}}-\varphi_{\eta}^{\bm{\mu}})}-\absolutevalue{s^{\bm{\mu}}}\cos{(\varphi_{\sigma}^{\bm{\mu}}-\varphi_{\eta}^{\bm{\mu}})}\geq
0,\quad\forall\eta.$ (230)

Therefore, an alternative way of satisfying the $\bm{\mu}$-component of the
full set of sufficient conditions, is to set the phases of $\tau$ such that
they cancel those of $\eta$

$\displaystyle\varphi_{\tau}^{\bm{\mu}}=\varphi_{\eta}^{\bm{\mu}},$ (231)

for all $\eta$. Substituting this phase choice on $\tau$ into the sufficient
condition in Eq. (230) gives

$\displaystyle
f^{\bm{\mu}}(\rho)\absolutevalue{t^{\bm{\mu}}}-\absolutevalue{s^{\bm{\mu}}}\cos{(\varphi_{\sigma}^{\bm{\mu}}-\varphi_{\eta}^{\bm{\mu}})}\geq
0,\quad\forall\eta.$ (232)

Noting that $\max_{x}\cos{x}=1$, we see that the above equation holds for all
$\eta$ if

$\displaystyle
f^{\bm{\mu}}(\rho)\absolutevalue{t^{\bm{\mu}}}\geq\absolutevalue{s^{\bm{\mu}}},$
(233)

which together with Eq. (231) gives us a second $\eta$-independent way of
satisfying the $\bm{\mu}$-component of Eq. (228). Putting back in our explicit
labels $\bm{\mu}=\\{\lambda,\alpha,j\\}$, we have that, if the $\tau$ we chose
at every $\eta$ was a valid quantum state such that

$\displaystyle
f^{\lambda}_{j}(\rho)t^{(\lambda,\alpha)}_{j}=s^{(\lambda,\alpha)}_{j}\,\text{
or
}\,f^{\lambda}_{j}(\rho)\absolutevalue{t^{(\lambda,\alpha)}_{j}}\geq\absolutevalue{s^{(\lambda,\alpha)}_{j}},\
\quad\forall\lambda,j,\alpha,$ (234)

then $\rho\xrightarrow{G}\sigma$, and thus Eq. (234) constitutes a sufficient
condition on the transition $\rho\xrightarrow{G}\sigma$, as claimed. ∎

We now present the following theorem from which Theorem 8 in the main text
follows as a corollary.

######  Theorem 31.

Assume without loss of generality (Lemma 14) that ${\cal G}(\sigma)$ is full-
rank. Then there exists a $G$–covariant operation transforming $\rho$ into
$\sigma$ if

$\displaystyle\lambda_{\mathrm{min}}n^{-1}f^{\lambda}_{j}(\rho)\geq
g^{\lambda}_{j}(\sigma),\quad\forall\lambda\neq 0,j,$ (235)

where we have defined
$f^{\lambda}_{j}(\rho)\coloneqq\mathrm{tr}[\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{2}}\rho^{\lambda^{*}}_{j}{\cal G}(\rho)^{-\frac{1}{2}}]$,
$g^{\lambda}_{j}(\sigma)\coloneqq\sum_{\alpha}\absolutevalue{\mathrm{tr}[X^{(\lambda,\alpha)\dagger}_{j}\sigma]}$,
$\lambda_{\mathrm{min}}$ as the smallest eigenvalue of ${\cal G}(\sigma)$, and
$n$ as the sum of the dimensions of all distinct non-trivial irreps appearing
in the representation of $G$ on ${\cal B}({\cal H}_{B})$, with ${\cal H}_{B}$
being the Hilbert space of the output system.

###### Proof.

We begin by making an explicit choice for $\\{\tau_{\Theta}\\}$ in Lemma 30
and proceed by showing that each $\tau_{\Theta}$ corresponds to a valid
quantum state.

For the $\lambda=0$ irrep we always choose

$\displaystyle t^{(0,\alpha)}_{0}\coloneqq s^{(0,\alpha)}_{0},\
\forall\alpha,$ (236)

which corresponds to setting ${\cal G}(\tau_{\Theta})={\cal G}(\sigma)$. Due
to (F3), this choice guarantees that all zero mode conditions from Lemma 30
are satisfied for any given $\sigma$.

For $\lambda\neq 0$, let us choose

$\displaystyle t^{(\lambda,\alpha)}_{j}\coloneqq
e^{i\theta^{(\lambda,\alpha)}_{j}}c^{\lambda}_{j}\absolutevalue{s^{(\lambda,\alpha)}_{j}},\quad
c^{\lambda}_{j}\coloneqq\frac{\lambda_{\min}}{n\left(\sum_{\alpha}\absolutevalue{s^{(\lambda,\alpha)}_{j}}\right)},\quad\theta^{(\lambda^{*},\alpha)}_{j}=-\theta^{(\lambda,\alpha)}_{j}$
(237)

where $n$ is the sum of the dimensions of all distinct non-trivial irreps
appearing in the representation of $G$ on ${\cal B}({\cal H}_{B})$.

If $\tau_{\Theta}$ corresponds to a valid quantum state for all values of
$\theta^{(\lambda,\alpha)}_{j}$, from Lemma 30 we immediately have the
sufficient condition

$\displaystyle f^{\lambda}_{j}(\rho)c^{\lambda}_{j}\geq 1,\forall\lambda\neq
0,j,$ (238)

which, with the identification
$g^{\lambda}_{j}(\sigma)\coloneqq\sum_{\alpha}\absolutevalue{s^{(\lambda,\alpha)}_{j}}=\sum_{\alpha}\absolutevalue{\mathrm{tr}[X^{(\lambda,\alpha)\dagger}_{j}\sigma]}$,
gives the statement of the lemma.

All that is left to do then is demonstrate that these choices of coefficients
$t^{(\lambda,\alpha)}_{j}$ indeed correspond to a valid quantum state
$\tau_{\Theta}$. We first note that the choice
$t^{(0,0)}_{0}=s^{(0,0)}_{0}=\frac{1}{\sqrt{d}}$ ensures that $\tau_{\Theta}$
has trace 1. Furthermore, since $\tau_{\Theta}={\cal
G}(\sigma)+\sum_{\begin{subarray}{c}\lambda,\alpha,j:\\\ \lambda\neq
0\end{subarray}}t^{(\lambda,\alpha)}_{j}X^{(\lambda,\alpha)}_{j}$, where the
$t^{(\lambda,\alpha)}_{j}$ in the sum have been so chosen that
$t^{(\lambda^{*},\alpha)}_{j}=\left(t^{(\lambda,\alpha)}_{j}\right)^{*}$,
$\tau_{\Theta}$ is the sum of two Hermitian operators and so is also
Hermitian. All that remains is to verify that $\tau_{\theta}$ is positive
semidefinite, i.e.,

$\displaystyle\bra{\psi}\tau_{\Theta}\ket{\psi}\geq 0,$ (239)

for any pure state $\ket{\psi}$ of the output system. The left hand side of
this expression can be lower bounded in the following way for any
$\theta^{(\lambda,\alpha)}_{j}\in[0,2\pi)$:

$\displaystyle\bra{\psi}\tau_{\Theta}\ket{\psi}$
$\displaystyle=\sum_{\lambda,\alpha,j}t^{(\lambda,\alpha)}_{j}\bra{\psi}X_{j}^{(\lambda,\alpha)}\ket{\psi}$
(240)
$\displaystyle=\sum_{\alpha:\lambda=0}s^{0,\alpha}_{0}\bra{\psi}X_{0}^{(0,\alpha)}\ket{\psi}+\sum_{\begin{subarray}{c}\lambda,\alpha,j:\\\
\lambda\neq
0\end{subarray}}e^{i\theta^{(\lambda,\alpha)}_{j}}c^{\lambda}_{j}\absolutevalue{s^{(\lambda,\alpha)}_{j}}\bra{\psi}X_{j}^{(\lambda,\alpha)}\ket{\psi}$
(241) $\displaystyle\geq\bra{\psi}{\cal
G}(\sigma)\ket{\psi}-\sum_{\begin{subarray}{c}\lambda,\alpha,j:\\\ \lambda\neq
0\end{subarray}}c^{\lambda}_{j}\absolutevalue{s^{(\lambda,\alpha)}_{j}}\absolutevalue{\bra{\psi}X_{j}^{(\lambda,\alpha)}\ket{\psi}}$
(242)
$\displaystyle\geq\lambda_{\min}-\sum_{\begin{subarray}{c}\lambda,\alpha,j:\\\
\lambda\neq
0\end{subarray}}c^{\lambda}_{j}\absolutevalue{s^{(\lambda,\alpha)}_{j}}$ (243)
$\displaystyle=\lambda_{\min}-\sum_{\begin{subarray}{c}\lambda,\alpha,j:\\\
\lambda\neq
0\end{subarray}}\frac{\lambda_{\min}}{n\left(\sum_{\alpha^{\prime}}\absolutevalue{s^{(\lambda,\alpha^{\prime})}_{j}}\right)}\absolutevalue{s^{(\lambda,{\alpha})}_{j}}$
(244) $\displaystyle=\lambda_{\min}-\sum_{\lambda\neq
0,j}\frac{\lambda_{\min}}{n}$ (245)
$\displaystyle=\lambda_{\min}-\lambda_{\min}=0,$ (246)

where in the second inequality we have used the fact that the operators
$\left\\{X^{(\lambda,\alpha)}_{j}\right\\}$ are normalized. This confirms that
$\tau_{\Theta}\geq 0$ for all $\theta^{(\lambda,\alpha)}_{j}\in[0,2\pi)$,
which completes the proof. ∎

The Sandwiched $\alpha$–Rényi divergence $D_{\alpha}(\rho||\sigma)$ for two
states $\rho,\sigma$ of a quantum system is defined as [61, 62]

$D_{\alpha}(\rho||\sigma):=\frac{1}{\alpha-1}\log\mathrm{tr}\left[\sigma^{\frac{1-\alpha}{2\alpha}}\rho\sigma^{\frac{1-\alpha}{2\alpha}}\right]^{\alpha},$
(247)

whenever the support of $\rho$ lies in the support of $\sigma$, and is
infinite otherwise.

In the case of $\alpha=2$, we extend the domain of the first argument to the
set of all linear operators in the support of $\sigma$ in the following
manner:

$D_{2}(X||\sigma):=\log\mathrm{tr}\left(\left[\sigma^{-\frac{1}{4}}X\sigma^{-\frac{1}{4}}\right]^{\dagger}\left[\sigma^{-\frac{1}{4}}X\sigma^{-\frac{1}{4}}\right]^{\dagger}\right).$
(248)

When $X$ is Hermitian, which is the case whenever $X$ is a valid quantum
state, this extension reduces to the standard definition of the sandwiched
$\alpha$–Réyni divergence for $\alpha=2$. We can then summarize the theorem
above as the following statement, reproduced from the main text:

See 8

###### Proof.

We consider the transition $\rho\xrightarrow{G}\sigma_{p}$, where
$\sigma_{p}\coloneqq(1-p)\sigma+p\frac{\mathbbm{1}}{d}$. We begin by noting
that $\lambda_{\mathrm{min}}[{\cal
G}(\sigma_{p})]=(1-p)\lambda_{\mathrm{min}}+\frac{p}{d}$ and
$g^{\lambda}_{j}(\sigma_{p})=(1-p)g^{\lambda}_{j}(\sigma)$ when
$(\lambda,j)\neq(0,0)$.

Therefore, substituting
$\sigma_{p}\coloneqq(1-p)\sigma+p\frac{\mathbbm{1}}{d}$ into Theorem 31 gives
the sufficient condition on the transition $\rho\xrightarrow{G}\sigma_{p}$:

$\displaystyle n^{-1}\left(\lambda_{\rm
min}+\frac{p}{d(1-p)}\right)f^{\lambda}_{j}(\rho)\geq
g^{\lambda}_{j}(\sigma),\forall(\lambda,j)\neq 0.$ (249)

We now express this in terms of the extended sandwiched $\alpha$–Rényi
divergence $D_{\alpha}(\rho||\sigma)$ for $\alpha=2$ defined in Eq. (248), and
it is readily checked that $D_{2}(\rho^{\lambda}_{j}||{\cal G}(\rho))=\log
f^{\lambda}_{j}(\rho)$ for all $\lambda,j$. Since $\log(x)$ is monotonic in
$x$, we can re-express Eq. (249) as in the statement of the theorem,
completing the proof. ∎

###  H.4 Conditions for identical input and output systems

When the input system $A$ and output system $B$ are identical, one can,
without loss of generality, replace ${\cal E}^{\eta}$ in Lemma 27 with
$(1-q)\mathcal{I}+q{\cal E}^{\eta}$, where $q$ is a probability and
$\mathcal{I}$ is the identity channel. The conditions of Lemma 27 are then
rewritten as:

$\displaystyle(1-q)\langle\eta,\rho\rangle+q\langle\eta,{\cal
E}^{\eta}(\rho)\rangle\geq\langle\eta,\sigma\rangle,\mbox{ for all }\eta\mbox{
and any }q\in[0,1],$ (250)

for an arbitrary family of covariant channels $\\{{\cal E}^{\eta}\\}$
parameterised by $\eta$, which can be rearranged as

$\displaystyle q\langle\eta,{\cal
E}^{\eta}(\rho)\rangle\geq\langle\eta,(\sigma-(1-q)\rho)\rangle,\mbox{ for all
}\eta\mbox{ and any }q\in(0,1],$ (251)

and so we define

$\sigma(q):=\sigma-(1-q)\rho,$ (252)

for any $q\in[0,1]$ for which we require

$\displaystyle q\langle\eta,{\cal
E}^{\eta}(\rho)\rangle\geq\langle\eta,\sigma(q)\rangle,\mbox{ for all
}\eta\mbox{ and any }q\in[0,1].$ (253)

Therefore, when the input and output systems are the same, Lemma 30 can be
upgraded to

######  Lemma 32.

Let us define components $t^{(\lambda,\alpha)}_{j}\coloneqq\langle
X^{(\lambda,\alpha)}_{j},\tau\rangle$ and
$s(q)^{(\lambda,\alpha)}_{j}\coloneqq\langle
X^{(\lambda,\alpha)}_{j},\sigma(q)\rangle$. When the input system $A$ and the
output system $B$ are identical, the state $\rho$ can be transformed into
$\sigma$ covariantly with respect to a symmetry group $G$ (i.e.
$\rho\xrightarrow{G}\sigma$) if there exists a family of valid quantum states

$\displaystyle\left\\{\forall\theta^{(\lambda,\alpha)}_{j}\in[0,2\pi):\tau_{\Theta}=\sum_{\lambda,\alpha,j}t^{(\lambda,\alpha)}_{j}X^{(\lambda,\alpha)}_{j}\right\\},$
(254)

where each $(\lambda,\alpha,j)$-component of $\tau_{\Theta}$ satisfies either:

$\displaystyle qf^{\lambda}_{j}(\rho)t^{(\lambda,\alpha)}_{j}$
$\displaystyle=s(q)^{(\lambda,\alpha)}_{j},\text{ or,}$ (255) $\displaystyle
qf^{\lambda}_{j}(\rho)\absolutevalue{t^{(\lambda,\alpha)}_{j}}-\absolutevalue{s(q)^{(\lambda,\alpha)}_{j}}$
$\displaystyle\geq 0,\quad
t^{(\lambda,\alpha)}_{j}=e^{i\theta^{(\lambda,\alpha)}_{j}}\absolutevalue{t^{(\lambda,\alpha)}_{j}}.$
(256)

for some probability $q\in[0,1]$.

###### Proof.

The proof follows that of Lemma 30 with ${\cal E}^{\eta}\rightarrow q{\cal
E}^{\eta}$ and $\sigma\rightarrow\sigma(q)$. ∎

These conditions are identical to those in Lemma 30 under
$f^{\lambda}_{j}(\rho)\rightarrow qf^{\lambda}_{j}(\rho)$ and
$\sigma\rightarrow\sigma(q)$. Making use of this lemma, we find that, when the
input and output systems are the same, Theorem 31 can be similarly upgraded to

######  Theorem 33.

Assume that ${\cal G}(\sigma)$ is full-rank. When the input and output systems
are identical, there exists a $G$–covariant operation transforming $\rho$ into
$\sigma$ if $\rho=\sigma$ or if there exists a probability $q\in(0,1]$ such
that

$\displaystyle\lambda_{\mathrm{min}}({\cal
G}[\sigma(q)])n^{-1}f^{\lambda}_{j}(\rho)\geq
g^{\lambda}_{j}(\sigma(q)),\quad\forall\lambda\neq 0,j,$ (257)

and ${\cal G}(\sigma(q))\geq 0$, where we have defined
$f^{\lambda}_{j}(\rho)\coloneqq\mathrm{tr}[\rho^{\lambda}_{j}{\cal
G}(\rho)^{-\frac{1}{2}}\rho^{\lambda^{*}}_{j}{\cal G}(\rho)^{-\frac{1}{2}}]$,
$g^{\lambda}_{j}(\sigma(q))\coloneqq\sum_{\alpha}\absolutevalue{\mathrm{tr}[X^{(\lambda,\alpha)\dagger}_{j}\sigma(q)]}$,
$n$ as the sum of the dimensions of all distinct non-trivial irreps appearing
in the representation of $G$ on ${\cal B}({\cal H}_{B})$, with ${\cal H}_{B}$
being the Hilbert space of the output system, and
$\lambda_{\mathrm{min}}({\cal G}[\sigma(q)])$ as the smallest eigenvalue of
${\cal G}(\sigma(q))$.

###### Proof.

The only covariantly accessible state at $q=0$ from $\rho$ is $\rho$ itself.
At all other values of $q$, the proof follows that of Theorem 31 with the
substitutions $f^{\lambda}_{j}(\rho)\rightarrow qf^{\lambda}_{j}(\rho)$ and
$\sigma\rightarrow\sigma(q)$, except for the following caveat. The part of
that proof demonstrating why $\tau_{\Theta}$ is positive relies on ${\cal
G}(\tau_{\Theta})$ being positive, which may not be true for all $q\neq 0$,
since we now assign ${\cal G}(\tau_{\Theta})\coloneqq q^{-1}{\cal
G}(\sigma(q))$, but because $q^{-1}\sigma(q)$ or equivalently ${\cal
G}(\sigma_{q})$ may not be positive for all $q\neq 0$, $q^{-1}{\cal
G}(\sigma(q))$ may not be either. We therefore additionally impose ${\cal
G}(\sigma(q))\geq 0$. ∎

By applying Theorem 33 to a partially depolarised version of $\sigma$,
$\sigma_{p}\coloneqq(1-p)\sigma+p\frac{\mathbbm{1}}{d}$, and noting once again
that $D_{2}(\rho^{\lambda}_{j}||{\cal G}(\rho))=\log f^{\lambda}_{j}(\rho)$,
we obtain Theorem 9 of the main text as an immediate corollary. We note that
the assumption of ${\cal G}(\sigma)$ being full-rank can be dropped by using
Lemma 14 to justify truncating the output Hilbert space to
$\operatorname{supp}({\cal G}(\sigma))$, and then redefining
$\sigma_{p}\rightarrow(1-p)\sigma+p\frac{\mathbbm{1}}{d_{S}}$ where $d_{S}$ is
the dimension of $\operatorname{supp}[{\cal G}(\sigma)]$,
$\lambda_{\mathrm{\min}}[\cdot]$ becomes $\min_{\ket{\psi}\in{\cal
H}_{S}}\bra{\psi}(\cdot)\ket{\psi}$, and $n$ becomes the sum of the dimensions
of all distinct non-trivial irreps appearing in the representation of ${\cal
G}$ on ${\cal B}(\operatorname{supp}[{\cal G}(\sigma)])$.