TITLE: $|G|>2$ implies $G$ has non trivial automorphism
QUESTION [52 upvotes]: Well, this is an exercise problem from Herstein which sounds difficult:

How does one prove that if $|G|>2$, then $G$ has non-trivial automorphism?

The only thing I know which connects a group with its automorphism is the theorem, $$G/Z(G) \cong \mathcal{I}(G)$$ where $\mathcal{I}(G)$ denotes the Inner- Automorphism group of $G$. So for a group with $Z(G)=(e)$, we can conclude that it has a non-trivial automorphism, but what about groups with center?

REPLY [6 votes]: The other two answers assume the axiom of choice:

Arturo Magidin uses choice when he forms the direct sum ("...it is isomorphic to a (possibly infinite) sum of copies of $C_2$...")
HJRW uses choice when he fixes a basis (the proof that every vector space has a basis requires the axiom of choice).

If we do not assume the axiom of choice then it is consistent that there exists a group $G$ of order greater than two such that $\operatorname{Aut}(G)$ is trivial. This is explained in this answer of Asaf Karagila.