Let $\mathcal{M}_2(\mathbb{C})$ be the set of square matrices of dimension 2, and $\mathcal{S}_2(\mathbb{C})$ the subset of self-adjoint matrices. The function $f_X: \mathcal{S}_2(\mathbb{C}) \rightarrow \mathcal{S}_2(\mathbb{C})$ is defined by $$ f_X(A) = \sum_{n=0}^\infty X^n A (X^\dagger)^n $$ where $X \in \mathcal{M}_2(\mathbb{C})$, and where $X^\dagger$ is the Hermitian adjoint of $X$. The problem is the following:
Is there a non-trivial matrix $X$ such that, $\forall A, B \in \mathcal{S}_2(\mathbb{C})$, if $\mathrm{Tr}\left[AB\right] = 0$, then $\mathrm{Tr}\left[f_X(A)f_X(B)\right] =0$?
Any hint would be appreciated!
EDIT: As mentionned in the comments, $X=0$ and $X = cI$ with $|c| < 1$ are obvious solutions. Also, if $X^d = cI$, then $f_X(A) = \sum_{n=0}^\infty c^n \sum_{m=0}^{d-1} X^m A (X^\dagger)^m = \frac{1}{1-c} \sum_{m=0}^{d-1} X^m A (X^\dagger)^m$. Furthermore, the eigenvalues $\lambda_i$ of $X$ are such that $\lambda_i^d = c$, so one can diagonalize $X = PDP^{-1}$ but I did not manage to go further when substituting in the trace.
