I had this "almost bonus" question on the final in Group Theory recently: prove that there is no such group $G$ which would satisfy ${\rm Aut}(G)\cong \mathbb{Z}_n$, where $n$ is an odd integer. I don't have much certainty if this proof is OK, and one's opinion would be appreciated.
Here's my attempt:
Suppose that there exists such a group $G$ that satisfies the above condition. Since $\mathbb{Z}_n$ is cyclic, ${\rm Aut}(G)$ is also cyclic and $G$ is abelian, which implies that ${\rm Inn}(G)\cong \{e\}$. Thus ${\rm Aut}(G)={\rm Out}(G)$. (I don't think this fact is important here though). Now, since an automorphism sends a generator to a generator, and since each automorphism is completely determined by such mapping, $|{\rm Aut}(G)|$ must be of factorial order. But an integer factorial is always even. However, $|\mathbb{Z}_n|$ is odd, which is a contradiction. Therefore, no such $G$ exists.
I'm afraid that my proof is not very nice, but at least a genuine attempt was made.
Best Answer
$\DeclareMathOperator{\Aut}{Aut}$ We'll show the slightly stronger result that $|\Aut(G)| \not= 2n+1$ for any $n \ge 1$ EDIT: Provided $G$ is abelian: see the comments.
If $\Aut(G)$ is trivial, then there is nothing to show. Suppose, then, that $G$ has a nontrivial automorphism group. We will show that $|\Aut(G)|$ is even by explicitly exhibiting an automorphism $\phi$ of $G$ having order $2$ for any group $G$. It will then follow by Lagrange's theorem that $\Aut(G)$ has even order.
If $G$ has an element of order at least $3$, then $\phi:x \to x^{-1}$ is the desired automorphism of even order. It only remains to consider the case where $G$ contains only elements of order less than or equal to $2$. In that case, we must have that $G$ is a vector space over $\mathbb{Z}_2$ (see Group where every element is order 2 for a proof of this result). If $G$ is one dimensional as a vector space, then $\Aut(G) = \Aut(\mathbb{Z}_2)$ is trivial, so there is nothing more to show. For $G$ having dimension greater than $1$, we have that $\Aut(G) \cong \operatorname{GL}(G)$, so if we pick an ordered basis for $G$, then the automorphism which interchanges the first and second basis vectors (and fixes all others) is an order $2$ automorphism. (This argument, which Slade gave in the comments, applies even to infinite $G$, while my previous argument tacitly assumed $G$ was finite.)