Examples of non-cyclic group with a cyclic automorphism group In introduction to algebra we got the exercise:


  Let $G$ be a group. Show that when $\operatorname{Aut}(G)$ is cyclic $G$ is abelian.


This doesn't make that much trouble. Denote the center (all commuting elements) with $Z$.
Then $G/Z$ is isomorphic to $\operatorname{Int}(G)$ where $\operatorname{Int}(G)$ denotes the subgroup of inner automorphisms. As every subgroup of a cyclic group is cyclic we have $G/Z$ is cyclic and hence $G$ is abelian (As a group which is cyclic of the center is abelian).

So the question is: 


  Is there a non cyclic group with a cyclic automorphism group?


We discussed the question already in chat and I found (thanks to google) a solution there, but I would enjoy other Examples.
Whenever $G$ is finite and its automorphismus is cyclic we can already conclude that $G$ is cyclic. 


Because as we already saw $G$ is abelian and finite, we can use the fundamental theorem of finitely generated abelian groups and say that wlog 
$G=\mathbb{Z}/p^k\mathbb{Z} \times \mathbb{Z}/p^j \mathbb{Z}$. But the automorphismgroup isn't abelian and hence isn't cyclic. 


  For non finite groups the implication isn't true.


The following is from this link and only slightly reworded.

Let $G$ be the subgroup of the additive group of rational numbers comprising those rational numbers that, when written in reduced form, have denominators that are square-free numbers, i.e., there is no prime number $p$  for which $p^2$  divides the denominator.

Then:
The only non-identity automorphism of  is the negation map, so the automorphism group is  $\mathbb{Z}/2\mathbb{Z}$, and is hence cyclic.

The group $G$ is not a cyclic group. In fact, it is not even a finitely generated group because any finite subset of  can only cover finitely many primes in their denominators. It is, however, a locally cyclic group: any finitely generated subgroup is cyclic.
$\mathbb{Q}^*$ is non cyclic, but the only automorphism is the identity.