Let $M$ be an infinite $R$-module, for a commutative ring $R$. Let $F \subseteq M$ be the set of elements which have a trivial annihilator: that is, for all $f \in F$, $$ rf = 0_M \iff r = 0_R.$$ Suppose that $T: M \to M$ preserves $F$: that is, $f \in F \implies Tf \in F$. Does it follow that $T$ is left-invertible? Does this depend on whether $R$ is finite?
Let R be an integral domain that is not a field and x be a nonzero element of R.
The map $T:r\mapsto rx$ maps F into F for the left R module R. If T were left invertible, that would mean there is a y such that $xy=1$. Since you can choose an element x that isn't a unit, it's possible for T to be noninvertible.
This same example does not survive if R is finite, though, since a finite domain is a field.
Here's a different example: let $M=\Bbb Z/(n)$ be viewed as a $\Bbb Z$ module. Then $F$, being empty, is obviously preserved by whatever module endomorphism you choose, and there are many noninvertible endomorphisms of this module. This one shows finiteness is not much help.