Note that because $A$ is size $n$ with $n$ distinct eigenvectors, $Av = \lambda_i v$ implies that $v$ is a multiple of $v_i$.
Thus, for every $i$, we note that $$ A(Mv_i) = MAv_i = \lambda_i (Mv_i) $$ which implies that $Mv_i$ is a multiple of $v_i$, which implies that $v_i$ is an eigenvector of $M$.