Automorphism Group isomorphic to $\mathbb{Z}_n$

Question

Suppose $n \geqslant 3$ such that $n$ is an odd integer, prove that there does not exist a group $G$ such that $\operatorname{Aut}(G)$ is isomorphic to $\mathbb{Z}_n$.

My idea is that if they are isomorphic, then they have the same number of elements of the same order. I try to find an element of order $2$ in $\operatorname{Aut}(G)$, then I can conclude such $G$ does not exist. However, I cannot find an element of order $2$ for $\operatorname{Aut}(G)$...

Answer 1

Can you show that any Abelian group of order $\ge3$ has an automorphism of order $2$?

If so, you can reduce to the non-Abelian case. If $G$ is non-Abelian, its inner automorphisms form a group isomorphic to $G/Z$ where $Z$ is its centre. But if $G/Z$ is cyclic, it's easy to prove that $G$ is Abelian, a contradiction.

Answer 2

Consider the automorphism $\alpha: x \to x^{-1}$. For any abelian group G it is an automorphism, and whenever $|G| > 2$, it has order $2$ (in $C_2$ and the trivial group it's simply the identity). Hence $\operatorname{Aut}G \not\cong \mathbb{Z}_n$ for any odd $n$.

Whenever $G$ is not abelian, use the tactic provided by Lord Shark the Unknown: the inner automorphisms of $G$ are a subgroup of $\mathbb{Z}_n$ and are hence cyclic. But $\operatorname{Inn}G \cong G/Z(G)$, so $G/Z(G)$ is cyclic, which implies that $G$ is abelian, a contradiction.