Q:

Action of ${\rm Aut}(G)$ on $G$

Let $G$ be a finite group and consider the natural action of ${\rm Aut}(G)$ on $G$  and let there are  two orbits under this action.
How could we show that 
$G$ is an (elementary) abelian group? Is the converse true?
(${\rm Aut}(G)$ denotes the automorphism group of $G$).
Thank you very much!
Let me add my motivation from this question. If we consider the action of ${\rm Inn}(G)$ on $G$ with two orbits, then we can show that $|G|=2$. Actually this question is a generalization of this property.

A:

We can easily assume $G$ is not trivial. By Lagrange's theorem, if $p$ divides $|G|$, then there exists $g \in G \backslash \{e\}$ with $g^p = 1$ (denote the identity of $G$ by $e$). Since the action of $\mathrm{Aut}(G)$ is transitive on $G \backslash \{e\}$ and that automorphisms preserve order, every non-trivial element of $G$ has order $p$. 
To show $G$ is abelian, note that since $G$ is a $p$-group, it has order $p^n$ for some $n$, hence its center is not trivial (this follows from the class equation : 
$$
|G| = |Z(G)| + \sum_{i=1}^r |G:C_G(g_i)|
$$
where $g_1,\cdots,g_r$ are representatives of the remaining non-trivial conjugacy classes of $G$). An automorphism must map the center to itself since if $g \in Z(G)$ and $h \in G$, then 
$$
\varphi(g) \varphi(h) = \varphi(gh) = \varphi(hg) = \varphi(h) \varphi(g). 
$$
Since the action of $\mathrm{Aut}(G)$ on $G \backslash \{e\}$ is transitive, we conclude that $G$ is abelian. 
Hope that helps,

