For $f\in \mathrm{End}_{\mathrm{End}_R (M)}(M)$ and $\{x_1,\ldots,x_n\}\subset M$ exists $r\in R$ such that $f(x_i)=r\cdot x_i, \, i=1,\ldots, n$.

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The exercise is as follows:

Let $R$ be an unitary ring (not necessarily commutative), $M$ a semisimple $R$-module.

Let $S=\mathrm{End}_R(M)$, so $M$ is an $S$-module with $T\cdot m=T(m)$, for $T\in S, m\in M$. Let $E=\mathrm{End}_S(M)$, $M$ is an $E$-module similarly.

(a) Prove that $N\subset M$ is an $R$-submodule of $M$ if and only if $N\subset M$ is an $E$-submodule of $M$.

(b) Prove that for all $f\in \mathrm{End}_S(M)$ and for all $\{x_1,\ldots,x_n\}\subset M$ exists $r\in R$ such that $f(x_i)=r\cdot x_i, \, i=1,\ldots, n$.

I managed to do (a) but I cannot prove (b).

I noted that if $x\in M$ and $f\in \mathrm{End}_S(M)$, as $Rx$ is an $R$-submodule of $M$, by part (a) $Rx$ is an $E$-submodule of $M$ then $f(Rx)\subset Rx$ therefore $f(x)=r\cdot x$ with $r\in R$. With this observation, for $x_i$ we can do the same, but how can I get the same $r$ for all $x_i$?

Can you suggest me any ideas to finish? Thanks in advance!