Automorphism group is transitive on non-identity elements implies characteristically simple

Verbal statement
If a group has the property that its automorphism group acts transitively on non-identity elements, then the group is characteristically simple.

Characteristically simple group
A characteristically simple group is a group whose automorphism group has no proper nontrivial characteristic subgroup.

Hands-on proof
Given: A group $$G$$, such that $$\operatorname{Aut}(G)$$ acts transitively on the set of non-identity elements of $$G$$. A characteristic subgroup $$H$$ of $$G$$

To prove: Either $$H$$ is trivial, or $$H = G$$

Proof: If $$H$$ is trivial, we are done. Otherwise, there exists a non-identity element $$x \in H$$. We want to show that $$H = G$$.

Pick any element $$y \in G$$. We want to argue that $$y \in H$$. Clearly, if $$y$$ is the identity element $$y \in H$$; otherwise, by assumption, there exists an automorphism $$\sigma \in \operatorname{Aut}(G)$$ such that $$\sigma(x) = y$$. But then, since $$H$$ is characteristic, and $$x \in H$$, we'd have $$y = \sigma(x) \in H$$. This completes the proof.

Conceptual proof
The automorphism group acting transitively, implies that if we start with any element of the group, its characteristic closure is the whole group -- this is precisely the condition of being characteristically simple.