Invariant superoperators under unitary action Suppose we work in a (finite) $d$-dimensional Hilbert space $ \mathcal H $ and let $ \mathcal B (\mathcal H) $ denote the set of bounded linear operators of this space. Define a superoperator $ \Phi : \mathcal B (\mathcal H) \to \mathcal B (\mathcal H) $ as a linear map between such bounded linear operators and let $ \mathcal S (\mathcal H ) $ denote the set of all such superoperators. In the below, we use $ U \in \mathcal B (\mathcal H) $ to denote a unitary operator in our space.
For vectors $ v \in \mathcal H $ we have the natural action $ v \mapsto U v $, which in turn induces an action $ f \mapsto U f U^{\dagger} := \mathcal U (f) $ due to the isomorphism $ \mathcal B (\mathcal H ) \simeq \mathcal H \otimes \mathcal H^{*} $(we use a calligraphic $ \mathcal U $ to denote the conjugation superoperator by the corresponding unitary $ U $).
On the level of superoperators, this action lifts to $ \Phi ( - ) \mapsto U^{\dagger} \Phi (U - U^{\dagger} ) U = \mathcal U^{\dagger} \circ \Phi \circ \mathcal U  $ (as before, we can view this as  arising from the natural isomorphism $ S( \mathcal H ) \simeq \mathcal B ( \mathcal H ) \otimes \mathcal B (\mathcal H)^{*} \simeq (\mathcal H \otimes \mathcal H^*) \otimes (\mathcal H \otimes \mathcal H^*) $).
My question is then what superoperators $ \Phi $ are invariant under this action for all unitaries, i.e. $ \forall \mathcal U, \; \mathcal U^{\dagger} \circ \Phi \circ \mathcal U = \Phi $? I've tried a few simple cases and it seems like the identity mapping $ \Phi(f) = f $ and the constant map $ \Phi(f) = I $ (where $ I $ is the identity operator) work, and hence any linear combination of the above but I am struggling to either find any more or prove these are the only ones.
Another idea I've noticed is that we could use view $ \Phi $ as a $ d^2 \times d^2 $ matrix and use vectorisation to transform the question into finding the invariant subspaces of the second tensor power, but in general it appears this question is difficult even if we know the invariant subspaces of the original space (for those familiar with the theory, the context of the question is related to finding trivial/irreducible representations of $\mathcal U (d) \otimes \mathcal U (d)$ but this isn't required to state the problem).
 A: Let $\Phi\in\mathcal{S}(\mathcal{H})$ and suppose that for any unitary
conjugation operator $\mathcal{U}\in\mathcal{S}(\mathcal{H})$,
$\mathcal{U}\Phi=\Phi\mathcal{U}$. Let $\{e_1,e_2,\dots,e_d\}$ be an
orthonormal basis for $\mathcal{H}$, let
$e_{ij}\in\mathcal{B}(\mathcal{H})$ denote the rank one operator
$e_i\otimes e_j$, the $d\times d$ matrix with $1$ in the $i$th row
$j$th column and zeros elswhere. Let $E_{ij}=\Phi(e_{ij})$. Claim,
there exists $r,s\in\mathbb{C}$ such that,
\begin{equation}
  \label{eq:wtf}
  E_{ij}=\begin{cases} re_{ij} & \text{ if } i\neq j, \text{ and }\\
    %
    re_{ij}+sI & \text{ if } i= j,
  \end{cases}
\end{equation}
so that for $B\in\mathcal{B}(\mathcal{H})$,
$\Phi(B)=rB+\mathrm{tr}(B)sI$. Accordingly, fix $1\leq i,j\leq d$ with
$i\neq j$, let $K_{ij}$ denote the span of $\{e_i,e_j\}$, and let
$P_{ij}$ denote the orthogonal projection onto $K_{ij}$. Notate the
compressions to $K_{ij}$ by
\begin{gather*}\epsilon_{ii}=P_{ij}E_{ii}\Bigm|_{K_{ij}}=
  \begin{bmatrix} a_{ii} & b_{ii} \\
    c_{ii} & d_{ii}  \end{bmatrix}\qquad
  %
  \epsilon_{ij}=P_{ij}E_{ij}\Bigm|_{K_{ij}}=
  \begin{bmatrix} a_{ij} & b_{ij} \\
    c_{ij} & d_{ij}  \end{bmatrix}\\
  %
  \epsilon_{ji}=P_{ij}E_{ji}\Bigm|_{K_{ij}}=
  \begin{bmatrix} a_{ji} & b_{ji} \\
    c_{ji} & d_{ji}  \end{bmatrix}\qquad
  %
  \epsilon_{jj}=P_{ij}E_{jj}\Bigm|_{K_{ij}}=
  \begin{bmatrix} a_{jj} & b_{jj} \\
    c_{jj} & d_{jj}  \end{bmatrix}
  %
\end{gather*}
Let $U_{1}$, $U_{2}$, and $U_{3}$ be the unitary matrices which fix
the orthogonal complement of $K_{ij}$ with action on $K_{ij}$ given by
$u_{1}=\begin{bmatrix} i & 0 \\ 0 & 1\end{bmatrix}$,
$u_{2}=\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$, and
$u_{3}=\frac{1}{\sqrt 2}\begin{bmatrix} 1 & 1 \\
  -1 & 1\end{bmatrix}$ respectively. Note that for
$A=\begin{bmatrix} a & b \\ c &
  d\end{bmatrix}\in\mathbb{C}\times\mathbb{C},$
$$u_1Au_1^\dagger=\begin{bmatrix} a & -ib \\ ic &  d\end{bmatrix}\quad
u_2Au_2^\dagger=\begin{bmatrix} d & c \\ b &  a\end{bmatrix}\quad
u_3Au_3^\dagger= \frac12\begin{bmatrix}
 a+b+c+d & -a+b-c+d \\ -a-b+c+d &  a-b-c+d \end{bmatrix}$$
For $k=1,2,3$, $U_kP_{ij}=P_{ij}U_k$ so that
$\mathcal{U}_{k}\Phi= \Phi\mathcal{U}_{k}$ implies for
$\ell,m\in\{i,j\}$ ,
\begin{equation}
  \label{eq:compression}
  u_k\epsilon_{\ell m}u_k^\dagger=
P_{ij}\Phi(U_ke_{\ell m}U_k^\dagger)\Bigm|_{K_{ij}}
\end{equation}
With $k=1$, equation (2) shows that the off-diagonal entries of
$\epsilon_{ii}$ and $\epsilon_{jj}$ equal zero and that all entries of
$\epsilon_{ij}$ and $\epsilon_{ji}$ except for $b_{ij}$ and $c_{ji}$
must equal zero. Since $\mathcal{U}_2(e_{ii})=e_{jj}$, $a_{ii}=d_{jj}$
and $d_{ii}=a_{jj}$. Since $\mathcal{U}_2(e_{ij})=e_{ji}$,
$b_{ij}=c_{ji}$. With these identities, it follows that
$2u_3\epsilon_{ij}u_3^\dagger=
\begin{bmatrix}b_{ij} & b_{ij} \\ -b_{ij} &
  -b_{ij} \end{bmatrix}$. Further, since
$2U_3e_{ij}U_3^\dagger=e_{ii}+e_{ij}-e_{ii}-e_{ii}$,
$$\begin{bmatrix}b_{ij} & b_{ij} \\ -b_{ij} &
  -b_{ij} \end{bmatrix}=
\begin{bmatrix} a_{ii}-d_{ii} & b_{ij} \\
  -b_{ij} & d_{ii}-a_{ii}\end{bmatrix},$$
so that $a_{ii}-d_{ii} = b_{ij}$. Letting $r=b_{ij}$ and  $s=d_{ii}$ one has,
$$\epsilon_{ii}=\begin{bmatrix} s+r & 0 \\
    0 & s  \end{bmatrix}\quad
  %
\epsilon_{ij}=\begin{bmatrix} 0 & r \\
    0 & 0  \end{bmatrix}\qquad
  %
\epsilon_{ji}=\begin{bmatrix} 0 & 0 \\
    r & 0  \end{bmatrix}\qquad
  %
\epsilon_{jj}=\begin{bmatrix} s & 0\\
    0 & s+r  \end{bmatrix}\qquad
  %
$$
Letting $i,j$ run through all unequal pairs yields equation (1).
Remark. Using similar techniques one can show the following.
Let $\Phi\in\mathcal{S}(\mathcal{H})$ and suppose that for any
orthogonal conjugation operator
$\mathcal{O}\in\mathcal{S}(\mathcal{H})$,
$\mathcal{O}\Phi=\Phi\mathcal{O}$. There exists $r,s,t\in\mathbb{C}$
such that, for $B\in\mathcal{B}(\mathcal{H})$,
$\Phi(B)=rB+sB^\top+\mathrm{tr}(B)tI$.
