Rings that are isomorphic to the endomorphism ring of their additive group. Every ring is isomorphic to a subring of the endomorphism ring of it's underlying group. That's Cayley's theorem for rings.
What can we say about rings that are isomorphic to the endomorphism ring of their underlying additive group? 
(are they always integral domains? ADDED: no, $Z/nZ$ is a counter example. from the comment by egreg)
 A: 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 $R \otimes S=0$.
