Let $A$ be a simple Artinian $K$-algebra with a minimal left ideal $M$. Here, $M$ can be viewed as a simple left $A$ module, and, by Schur's lemma, $D=\text{End}_A(M)$ is a $K$-algebra. By Wedderburn-Artin theorem, for any primitive idempotent $e$ of $A$, we have $Ae \cong M$ as modules and $eAe \cong D^{op}$ as $K$-algebras.
Suppose $\rho: A \to \text{End}_K(M^n)$ is a linear representation of $A$. For any $v \in M^n$, $av:=\rho(a)v$. Consequently, $$\text{End}_A(M^n)=\{\tau \in \text{End}_A(M^n) \mid \rho(a) \tau=\tau \rho(a) \ \ \forall a \in A\}$$
I want to simplify $\text{End}_A(M^n)$. What is the smallest subset $S$ of $A$ such that $$\text{End}_A(M^n)=\{\tau \in \text{End}_A(M^n) \mid \rho(a) \tau=\tau \rho(a) \ \ \forall a \in S\}?$$
Can $S=\{e\}$ for any primitive idempotent of $A$?
