Let $M$ be a noetherian $R$-module and let $f$ be a non-zero endomorphism of $M$ such that $\frac{M}{\ker(f)}\cong M$. I want to show that $f$ is a monomorphism. In order to do it so, I want to use the following Lemma.
Lemma If $M$ is a noetherian module and $f$ is a non-zero endomorphism of $M$, then $f$ is an automorphism if and only if it is an epimorphism.
So I want to prove that $f$ is a surjective function using the fact that $\frac{M}{\ker(f)}\cong M$ but I can't think of a way to prove it nor have I found a counterexample.
Composing the isomorphism $M/ker(f) \cong M$ with the quotient map gives a surjective endomorphism, and hence an automorphism. This means that $ker(f)$ must be trivial, so $f$ must be injective.
In fact, the above proof works for any Hopfian module $M$, and Noetherian modules are Hopfian. But $f$ need not be surjective, because a Noetherian module need not be co-Hopfian. For example, multiplication by $2$ is an injective but non-surjective endomorphism of $\mathbb{Z}$.