When is $G\cong\operatorname{End}(G)$? $\newcommand\End{\operatorname{End}}$Let $G$ be an Abelian group.
Are there sufficient conditions for the existence of an isomorphism $G\cong\End(G)$, where $\End(G)$ is considered a group under addition?
This is true for $\mathbb Z$ and $\mathbb Q$, but not (as far as I know) for $\mathbb R$, where you can construct monstrous additive functions from the Axiom of Choice.
Side questions for people to consider: If $R$ is a ring, when will it be isomorphic (as a ring) to the ring $\End_{\text{group}}(R)$ of group endomorphisms? When will it be isomorphic to the ring $\End_{\text{ring}}(R)$ of ring endomorphisms?
 A: A few partial answers:
1) if $G$ is a finitely generated abelian group, then $G\simeq\mathrm{End}(G)$ iff $G$ is cyclic.
indeed, if $G$ is infinite and has an element of prime order $p$, then $\mathrm{End}(G)$ has more elements of order $p$ than $G$. A similar counting argument also works when $G$ is finite and non-cyclic. And if $G$ is free abelian of rank $r\ge 2$, then $\mathrm{End}(G)$ is free abelian of rank $r^2$.
2) if $G$ is a subring of $\mathbf{Q}$ (e.g., $\mathbf{Z}[1/n]$ or $\mathbf{Q}$ itself), then $G\simeq\mathrm{End}(G)$. On the other hand, there are subgroups of $\mathbf{Q}$ such that $G$ is not isomorphic to $\mathrm{End}(G)$: for instance the subgroup of $\mathbf{Q}$ generated by the $1/p$ for prime $p$ has $\mathrm{End}(G)\simeq\mathbf{Z}$.
3) there are some more examples: for instance if $A$ is a subring of $\mathbf{Q}$ containing $1/n$ then $G=A\times(\mathbf{Z}/n\mathbf{Z})$ satisfies $G\simeq\mathrm{End}(G)$.
4) if $G\simeq H^{(X)}$ for some nonzero countable group $H$ and infinite set $X$ (this is the set of finitely supported functions $X\to H$), then $G$ is not isomorphic to $\mathrm{End}(G)$, just because $\mathrm{End}(G)$ has cardinal greater than $G$: indeed the cardinal of $G$ is that of $X$, while the cardinal of $\mathrm{End}(G)$ is that of $2^X$ (because $\mathrm{End}(G)$ contains $\mathrm{End}(H)^X$).
