TITLE: Existence of group $G$ with $|G|>2$ and automorphism group of odd order
QUESTION [8 upvotes]: In Rotman's introduction to the theory of groups, there is an exercise (7.9) where one has to prove that the following facts for a finite group $G$. 

If $G$ is abelian and $|G|>2$ then $\text{Aut}(G)$ has even order
$\text{Aut}(G)$ is not cyclic when $G$ is not abelian
$\text{Aut}(G)$ is never cyclic of odd order >1. 

This is all fairly easy, but I was wondering if there even exists a group $G$ with $|G|>2$ which has automorphism group of odd order. Anyone has an example or reference (couldn't find it by simple internet search)?
Clearly $G$ can't be abelian, symmetric, alternating, dihedral, ... so the most obvious counterexamples don't work.
Thanks in advance!

REPLY [2 votes]: Actually, it is currently known that the least nontrivial odd order automorphism group has order $3^7$. It was proved by Peter Hegarty and Desmond Machale in Minimal odd order automorphism groups
Another example is definitely not minimal, but still interesting enough to be mentioned. It is a complete group of order $3\cdot 7^{12}\cdot 19$, that was constructed by R.S. Dark in "A complete group of odd order" (here is the post, from which I found out about its existence: Does every finite non-trivial complete group have even order?). It also satisfies your condition as any complete group is isomorphic to its automorphism group and moreover it serves as a counterexample to a conjecture weaker than yours - the so-called Rose conjecture, that stated, that "all nontrivial finite complete groups have even order".