Does there exist a nontrivial abelian group $A$, such that $\mathrm{Aut}(A) \cong A$? Here $\mathrm{Aut}$ stands for the automorphism group.
What do I currently know:
If such $A$ exists, it has infinite exponent.
It follows from the first Prüfer theorem, that every bounded exponent noncyclic abelian group has non-abelian automorphism group. A cyclic group is isomorphic to its automorphism group iff it is trivial.
If such $A$ exists, it is infinitely generated.
It follows from the classification of finitely-generated abelian groups, that every finitely generated noncyclic abelian group has non-abelian automorphism group. A cyclic group is isomorphic to its automorphism group iff it is trivial.