Automorphism group of non-abelian group- Abelian or Non-Abelian?

I was asked this question in an interview-

Let G be an non-abelian group then what about it's automorphism group? Is it abelian or non-abelian?

I could not think about a single example at that time to contradict the proposition nor I know any general result to support it. Now I am thinking over it but still I am unable to find a suitable argument. Can anybody help me to solve this? Apology if this is too trivial. Thanks.

• What is the automorphism group of $S_3$? – Mathmo123 Aug 24 '16 at 11:51

Conjugations are part of the automorphism group, and are relatively easy to study. Given $a, b \in G$, we look at the automorphisms $g\mapsto aga^{-1}$ and $g\mapsto bgb^{-1}$.
Assume that the two automorphisms do commute. That means that for any $g\in G$, we have $abgb^{-1}a^{-1} = baga^{-1}b^{-1}$. Specifically, this holds for $g = ab$. We get $$ababb^{-1}a^{-1} = baaba^{-1}b^{-1}\\ ab = ba^2ba^{-1}b^{-1}\\ ab^2a = ba^2b$$ or, in other words, $ab$ commutes with $ba$. So, if you want to be certain that you have a non-abelian automorphism group, all you have to do is find a group with two elements $a, b$ such that $ab$ and $ba$ are unequal, and also do not commute.
Example: In $S_3$, the symmetric group on three objects, take $a = (1\,2\,3)$ and $b=(1\,2)$. In that case, $ab = (1\,3)$ while $ba = (2\,3)$. This makes $abba = (1\,3\,2)$ and $baab = (1\,2\,3)$. Therefore, the automorphisms given by conjugation by $a$ and by conjugation by $b$ do not commute.