Q:

${\rm Aut}(G)$ is cyclic $\implies G$ is abelian

I would appreciate if you could please express your opinion about my proof. I'm not yet very good with automorphisms, so I'm trying to make sure my proofs are OK.
Proof:
Since ${\rm Aut}(G)$ is cyclic, ${\rm Aut}(G)$ is abelian. Thus for any elements $\phi, \psi \in{\rm Aut}(G)$ and some elements $g_i \in G$, $\phi\psi(g_1g_2)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4=\phi\psi(g_2)\phi\psi(g_1)=\phi\psi(g_2g_1)=g_4g_3$. Hence, $g_3g_4=g_4g_3\implies G$ is abelian.
The proof seems to be quite straightforward, but I'd rather ask for advice.

A:

There is a nice chain of small results which proves this which continues down the path that Groups suggests.  If ${\rm Aut}(G)$ is cyclic, then so is any subgroup of it, in particular ${\rm Inn}(G)$.  ${\rm Inn}(G)\cong G/Z(G)$ where $Z(G)$ is the center.  If $G/Z(G)$ is cyclic, the group is abelian.

A:

The $\phi,\psi$ commute, and also the following steps are also OK:
$$\phi\psi(g_1g_1)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4.$$
Its not clear in your argument why $g_3g_4=\phi\psi(g_2)\phi\psi(g_1)$?
We are allowed to use commutativity of maps $\phi,\psi$, and we have to conclude commutativity of $g_1,g_2$. You may proceed in following directions.
(1) Consider a specific subgroup of ${\rm Aut}(G)$, namely ${\rm Inn}(G)$. How it is related with $G$?
(2) ${\rm Aut}(G)$ is cyclic, so is ${\rm Inn}(G)$, then using (1), what this will imply?

