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 $\tau$.

$\mathcal{C}$ is simply the span of the two maps that you noted (the identity and the trace) -- there is nothing else in the commutant.

One (admittedly somewhat roundabout) way of seeing this is to notice that if you unpack $\phi$ into an $n^2 \times n^2$ matrix $\Phi$ in the "usual" way (i.e., instead of thinking of it as a linear transformation acting on matrices, think of it as a matrix acting on their vectorizations), then your commutation relation is equivalent to $$ (U \otimes \overline{U})\Phi(U \otimes \overline{U})^* = \Phi $$ for all unitary $U \in \mathbb{C}^{n\times n}$ (here $\overline{U}$ is the entrywise complex conjugate of $U$).

This is the defining property of something called an isotropic state from quantum information theory, and it is well-known (see this paper, for example) that all matrices with this property are linear combinations of the identity matrix and the "maximally entangled state" $\rho = \sum_{i,j=1}^n \mathbf{e}_i\mathbf{e}_j^* \otimes \mathbf{e}_i\mathbf{e}_j^*$ (where $\{\mathbf{e}_i\}$ is the standard basis of $\mathbb{C}^n$). These two matrices correspond to the trace linear map and the identity linear map, respectively, once you "un-vectorize" everything.