Let $d\ge 2$ denote an integer, let $S_d$ denote the group of permutations of $d$ elements, and let $L(\mathbf{C}^d)$ denote the space of linear operators on the Hilbert space of $d$-dimensional complex vectors. For any $\pi\in S_d$ let us define the permutation unitary $U_\pi:= \sum_{i=0}^{d-1}|\pi(i)\rangle\langle i|$ (in bra-ket notation).
Let us define
$\text{Comm}(d):=\{ A\in L(\mathbf{C}^d\otimes \mathbf{C}^d) \,: [A,U_\pi\otimes U_\pi]=0\, \forall \,\pi\in S_d\}$.
What is a basis for the space $\text{Comm}(d)$?
I think that one can approach the problem with representation theory tools. Thank you for your time!
Here is a way of writing this for a representation theorist, and a start at the solution. Let $S_d$ be the symmetric group of degree $d$ acting on the $d$-element set $\Omega$. Let $\def\C{\mathbf{C}}$$V=\C[\Omega]$ be the standard $d$-dimensional complex representation of $S_d$. What is $\mathrm{Hom}_{S_d}(V\otimes V, V \otimes V)$?
Note that $V \otimes V \cong \C[\Omega^2] \cong V \oplus \C[V^{(2)}]$, where $V^{(2)} = \{(x, y) \in \Omega^2 : x \ne y\}$, and $\C[V^{(2)}] \cong M_{(d-2,1,1)}$, a Young permutation module. The decomposition of Young permutation modules into irreducible representations is described by the Kostka numbers. The result is that $$M_{(d-2,1,1)} \cong V_{(d)} \oplus V_{(d-1,1)}^2 \oplus V_{(d-2,2)} \oplus V_{(d-2,1,1)}.$$ Here I am assuming $d \ge 4$. Since $V \cong V_{(d)} \oplus V_{(d-1,1)}$ it follows that $$V\otimes V \cong V_{(d)}^2 \oplus V_{(d-1,1)}^3 \oplus V_{(d-2,2)} \oplus V_{(d-2,1,1)}.$$ Now apply Schur's lemma. The result is that $\dim \mathrm{Hom}_{S_d}(V\otimes V, V\otimes V) = 2^2 + 3^2 + 1 + 1 = 15$ for $d \ge 4$.
Another approach: Let $\chi$ be the character of $V$. Then $$\dim\mathrm{Hom}_{S_d}(V^{\otimes 2}, V^{\otimes 2}) = \langle \chi^2, \chi^2\rangle = \langle \chi^4, 1\rangle,$$ and by Burnside's lemma this is the number of orbits of $S_d$ on $\Omega^4$. Each orbit corresponds to a partition of $4$ into at most $d$ parts. There is $1$ orbit of type $4$, $4$ of type $3+1$, $3$ of type $2+2$, $6$ of type $2+1+1$ ($d \ge 3$), and $1$ of type $1+1+1+1$ ($d \ge 4$), so the answer is $1+4+3+6+1 = 15$ for $d \ge 4$.
More work is required to actually write down a basis for $\mathrm{Hom}_{S_d}(V\otimes V, V\otimes V)$, but it is possible in principle and should be somewhat mechanical. For a start, there are $4$ basis elements arising from $\mathrm{Hom}_{S_d}(V_{(d)}^2, V_{(d)}^2)$. These are $4$ rank-one operators of the form $v \otimes \phi$, where $v$ is either $\sum_i e_i \otimes e_i$ or $\sum_{i \ne j} e_i \otimes e_j$ and $\phi$ is either $\sum_i e_i^* \otimes e_i^*$ or $\sum_{i \ne j} e_i^* \otimes e_j^*$.