For finite group $G$ when is $|Aut(G)| < |G|$?

If $G$ is a finite group then we know $|Aut(G)|$ divides $(|G|-1)!$ ; I want to ask , if $G$ is a finite group with more than one element then is it true that $|Aut(G)| < |G|$ ( I know it is always true for cyclic groups) ? If $|Aut(G)| < |G|$ is not always true then can we characterize those groups for which it is true ?

• Well this is certainly true for all cyclic groups other than the trivial one because $\phi(n)<n$ for all $n\geq 1$. – RougeSegwayUser Oct 30 '14 at 4:28
• The automorphism group of the quaternion group is $S_4$, so it fails for this. – Peter Huxford Oct 30 '14 at 4:32
• Also, this is not the case for Alb. For this category, the problem amounts to enumerating the set of all $\mathbb{Z}$-bases. Even for $\mathbb{Z}_3^2$, this fails. It holds for $\mathbb{Z}_2 \times \mathbb{Z}_n$. In that case the number of $\mathbb{Z}$-bases is $\phi(n)$. Here I'm using Alb for the category of finite abelian groups (or $\mathbb{Z}$-modules). I'm fairly certain that this fails for all other finite abelian groups – RougeSegwayUser Oct 30 '14 at 4:42
• The automorphism group of $\Bbb{Z}_2\times\Bbb{Z}_2$ has order six. The automorphism group of $\Bbb{Z}_2^3$ has order 168... – Jyrki Lahtonen Oct 30 '14 at 5:02
• If the centre $Z(G) = 1$, then $G \cong \text{Inn}(G) \le \text{Aut}(G)$, so $|Aut(G)| \ge |G|$ in this case. e.g. all finite simple groups. – Alastair Litterick Oct 30 '14 at 20:17