Every element of G has the same order (either finite or infinite). G has no proper non-identity characteristic subgroups (it is characteristically simple), so considering the derived subgroup, G is either abelian or perfect. If G is finite, then it must be a p-group. Since a finite non-identity p-group is not perfect (maximal subgroups are normal), G is an elementary abelian p-group.
In general, if G is abelian, it must either be an elementary abelian p-group (abelian and all elements of the same finite order), or it must be a torsion-free divisible group (a vector space over $\mathbb{Q}$; choose f so that $f(nx) = x$, but then $n f(x) = f(nx) = x$ so x is divisible).
However, there are characteristically simple infinite p-groups, and I don't know that they do not have transitive automorphism groups. Three important examples are McLain groups (locally finite, lots of subnormals, upper triangular matrices over GF(p)), Hall groups (locally finite, few subnormals, wreath products), and Tarski monsters (2-generated simple groups).
As Steve mentions, in the torsion-free case, there are many non-abelian examples due to Higman (and Higman–Neumann–Neumann), since every torsion-free group embeds into a torsion-free group in which the inner automorphism group acts transitively on the non-identity elements.