Homogeneous groups
$G$ does not have to be abelian. There exist examples of infinite (necessarily highly nonabelian) groups with exactly two conjugacy classes; that is, even the action of inner automorphisms on non-identity elements is transitive.
As observed in the comments, since $[G, G]$ is a characteristic subgroup, if $G$ is nonabelian then it is necessarily perfect.
As observed in the comments, nonabelian free groups are not perfect.
I don't know the answer to the last question. It seems potentially difficult due to the existence of Tarski monsters.
Groupprops claims that for abelian groups this property is equivalent to being the underlying additive group of a field, although I don't immediately see a proof. As observed in the comments, as groups these are just vector spaces over $\mathbb{F}_p$ or $\mathbb{Q}$.
Edit: Here's a proof that if $G$ is abelian then it's a vector space over $\mathbb{F}_p$ or $\mathbb{Q}$. As you observed, if $G$ has an element of finite order then every element has order $p$ for some prime $p$, hence $G$ is an $\mathbb{F}_p$-vector space. Otherwise, every element has infinite order. If $g \in G$ and $n \in \mathbb{N}$, then by assumption there is an automorphism $\varphi : G \to G$ such that $\varphi(ng) = n \varphi(g) = g$. Hence $G$ is divisible, and an abelian torsion-free divisible group is a $\mathbb{Q}$-vector space.
You may be interested in this paper. The author proves that, in a certain sense, free groups of finite rank are homogeneous. The "certain sense" is, loosely, that if $a$ and $b$ satisfy the same formulae then they are connected by an automorphism.