If $K$ is a field of characteristic $0$, then the ring $KS_r$ is semisimple. Let $V$ be a $KS_r$-module. In particular it is semisimple. Can we assure that $\mathrm{End}_{KS_r}(V)$ is a semisimple ring?
Since $V$ is semisimple, it is a von Neumann regular ring, but this is not enough.
I presume $S_r$ is the symmetric group on $r$ symbols? If $G$ is any finite group, $K$ a field of characteristic zero, and $V$ is an irreducible module then $\text{End}_{KG}(V)$ is a skew field (Schur's lemma). More generally, if $V$ is finitely generated, it is a finite direct sums of irreducibles and $\text{End}_{KG}(V)$ is a direct product of matrix rings over skew fields so semisimple.
If $V$ is not finitely generated, then its endomorphism ring is too large to be semisimple.