Cyclic Automorphism group 
Show that no group can have its automorphism group cyclic of odd order.

I have shown it only if $G$ is cyclic, but I could not do that if $G$ is not cyclic. Can you help?
 A: Here are some hints:
Let $G$ be a group with a cyclic and odd group of automorphisms.


*

*Since $G/Z$ (Z being the center) is a subgroup of $\operatorname{Aut} G$, it will be cyclic. Deduce that $G = Z$, i.e. $G$ is abelian.

*Since $G$ is abelian, what can you say about $x\mapsto x^{-1}$ ? Deduce that $G \cong \bigoplus_{i} \mathbb Z/2\mathbb Z$, i.e. $G$ is an elementary abelian $2$-group.

*Now you know a lot about $G$, so you may try to find automorphisms of $G$ that will contradict that $\operatorname{Aut} G$ is cyclic.


I can give more hints if you tell where you're stuck.
-- editted:  for step 3: For instance: If there are at least 3 factors involved in the direct product $G \cong \bigoplus_i (\mathbb Z/2\mathbb Z)$ then permuting these factors gives rise to an automorphism (for instance $(a,b,c,...)\mapsto (b,a,c,...)$). This implies $S_3$ appears as subgroup of the automorphism-group so it will surely not be cyclic. If there are $2$ factors $G\cong (\mathbb Z/2\mathbb Z)^2$ and it's easy to see that any permutation of the three involutions of this group is an automorphism.
I think there is a more beautiful way to derive the contradiction but I don't see it right now.
-- editted (much later): I just thought of the more beautiful way: If the direct sum has at least two terms, consider the automorphism that switches these terms $(a,b,c,\dots)\mapsto (b,a,c,\dots)$. This is an automorphism of order 2, a contradiction.
