On the converse of Schur's Lemma

Solution 1:

There are counterexamples in general.

Suppose $V$ is a non-split extension of two non-isomorphic simple modules. I.e., $V$ has a unique simple submodule $W$ with simple quotient $U=V/W$ where $U\not\cong W$. Then any non-zero endomorphism of $V$ is an isomorphism, since the only possible kernel is $W$, but the image can't be $V/W=U$ since $V$ has no submodule isomorphic to $U$. If, further, $U$ and $W$ have $F$ as endomorphism ring, this isomorphism must be a scalar multiple of the identity.

For an explicit example, let $G$ be the group of invertible upper triangular $2\times 2$ matrices over a finite field $F$ with $\vert F\vert>2$, and let $V$ be the natural $2$-dimensional $FG$-module.