Rings that are isomorphic to the endomorphism ring of their additive group.

The question for which rings $R$ there is some isomorphism $R \cong \mathrm{End}_{\mathbb{Z}}(U(R))$ (where $U(R)$ denotes the underlying $\mathbb{Z}$-module) is rather unnatural and I bet that we cannot answer it.

It is more natural to ask for which rings $R$ the canonical homomorphism $\lambda : R \to \mathrm{End}_{\mathbb{Z}}(U(R))$ which maps $r \in R$ to $(x \mapsto rx)$ is an isomorphism. I will try to answer this.

Note that there is a homomorphism $\rho : U(\mathrm{End}_{\mathbb{Z}}(U(R))) \to U(R)$ mapping $f \mapsto f(1)$ (if these $U$'s confuse you, just forget about them; I am just saying that $\rho$ is additive, not multiplicative in general), and we have $\rho \circ U(\lambda) = \mathrm{id}_{U(R)}$. Thus, $\lambda$ is a monomorphism, and it is an isomorphism if and only if $U(\lambda) \circ \rho = \mathrm{id}$. And this means, by definition: If $\alpha : U(R) \to U(R)$ is a homomorphism, then $\alpha(x)=\alpha(1) x$ for all $x \in R$. In other words, $\alpha$ is $R$-linear (where $R$ acts on the right).

This is easily seen to be equivalent to the following: If $M$ is some right $R$-module, then any $\mathbb{Z}$-linear map $M \to R$ (I ignore the $U$s here) is automatically $R$-linear. Now let us look at a stronger property: If $M,N$ are any two right $R$-modules, then any $\mathbb{Z}$-linear map $M \to N$ is already $R$-linear. It turns out (by formal nonsense) that this property is equivalent to the condition that $\mathbb{Z} \to R$ is an epimorphism in the category of rings. One can classify these rings in the commutative case - Torsten Schöneberg has given a nice overview here:

If $R$ is a commutative ring, and $\mathbb{Z} \to R$ is an epi which is not injective, then $R=\mathbb{Z}/n$ for some $n \in \mathbb{Z}$. If it is injective, then there is a set of primes $\widehat{P}$, a subset $P$ and a function $n : P \to \mathbb{N}^+$, such that $R$ is isomorphic to the (possibly infinite) tensor product of the $(\widehat{P} \setminus P)^{-1}$-algebras $\mathbb{Z}/p^{n(p)} \times p^{-1} \mathbb{Z}$, where $p \in P$. Thus, the epimorphisms are built up canonically out of a) quotients, b) localizations, c) tensor products, d) directs products $R \times S$ with $\mathbb{Z} \to R,\,S$ epi such that $R \otimes S=0$.