Let $A$ be noetherian commutative ring with one, and let $f:A\to A$ be an epimorphic endomorphism of $A$.
Is $f$ necessarily an isomorphism?
("An epimorphic endomorphism" means of course "an endomorphism which is an epimorphism". In this post "ring" means "commutative ring with one", and morphisms are required to map $1$ to $1$.)
A few reminders:
A morphism of rings $f:A\to B$ is an epimorphism if for all pairs of morphisms $(g,h):B\rightrightarrows C$ the equality $g\circ f=h\circ f$ implies $g=h$. Surjective morphisms are epimorphic, but the converse does not hold: for instance the inclusion $\mathbb Z\to\mathbb Q$ is an epimorphism.
If $A$ is noetherian and $f:A\to A$ is a surjective endomorphism, then $f$ is an isomorphism, because if $f$ were not injective, then the kernels of the iterated endomorphisms $f^n$ would form an ascending chain of ideals of $A$.
If $A$ is a nonzero ring, then the map $(a_1,a_2,\dots)\to(a_2,a_3,\dots)$ is a surjective endomorphism of $B:=A\times A\times\cdots$ which is not an isomorphism (but of course $B$ is not noetherian).
A few links:
MathOverflow thread What do epimorphisms of (commutative) rings look like?.
Stacks Project Section Epimorphisms of rings.
Samuel Seminar . See in particular Section 2 of Exposé Number 7 by Daniel Ferrand.