Let $\mathcal{L}(\mathbb{C}^{n \times n})$ denote the algebra of all linear mappings from $\mathbb{C}^{n \times n}$ to $\mathbb{C}^{n \times n}$ and let $\mathcal{C} \subseteq \mathcal{L}(\mathbb{C}^{n \times n})$ denote the subalgebra of all $\phi \in \mathcal{L}(\mathbb{C}^{n \times n})$ which satisfy
$$
\phi(U^\*AU) = U^\*\phi(A)U
$$
for all $A \in \mathbb{C}^{n \times n}$ and all unitary $U \in \mathbb{C}^{n \times n}$ (in other words, $\mathcal{C}$ is the commutant of the set of all conjugations by unitary matrices).


**Question.** Is there an explicit description of $\mathcal{C}$?


Of course, there is some freedom for interpretation of the word "explicit"; I would be most happy with a set of mappings in $\mathcal{L}(\mathbb{C}^{n \times n})$ which spans $\mathcal{C}$.


**Remarks:**


* Clearly, the identity $\operatorname{id}\_{\mathbb{C}^{n \times n}}$ is an element of $\mathcal{C}$.
* The operator $\tau: \mathbb{C}^{n \times n}\to \mathbb{C}^{n \times n}$ given by 
$$
\tau(A) = \operatorname{tr}(A) \cdot \operatorname{id}\_{\mathbb{C}^{n \times n}}
$$
is an element of $\mathcal{C}$ (where $\operatorname{tr}(A)$ denotes the trace of the matrix $A$).
* The span of $\operatorname{id}\_{\mathbb{C}^{n \times n}} $ and $\tau$ is a subalgebra of $\mathcal{C}$ (since $\tau^2 = n\tau$), but I don't know whether $\mathcal{C}$ is larger than this span.



Commutant of the conjugations by unitary matrices
Building up on my comment, I can now give the complete answer. The space of matrices can be decomposed as follows:
$$
\mathbb M\_n(\mathbb C) = \mathbb C\cdot\mathrm{id}\oplus \mathfrak{sl}(n),
$$
where
$$
\mathfrak{sl}(n) = \{X\in\mathbb M\_n(\mathbb C)\mid \mathrm{Tr}(X) = 0\}.
$$


Thus, the conjugation representation of $\mathrm{U}(n)$ decomposes as the sum of a trivial representation and the conjugation repesentation on $\mathfrak{sl}(n)$. The latter is irreducible as a complex representation of $\mathrm{U}(n)$ because:


* the complexification of $\mathfrak{u}(n)$ is $\mathfrak{gl}(n)$, and
* the Lie algebra $\mathfrak{sl}(n)$ is a simple complex Lie algebra.


Therefore the algebra of linear $\mathrm U(n)$-equivariant maps is isomorphic to $\mathbb C\oplus \mathbb C$. The elements $(1,0)$ and $(0,1)$ are just orthogonal projections to $\mathbb C\cdot \mathrm{id}$ and its orthogonal complement $\mathfrak{sl}(n)$.


So, the space $\mathcal C$ from the question is indeed spanned by $\mathrm{id}\_{\mathbb M\_n}$ and $\tau$.


