let $G$ be a finite group. If $g$ is an automorphism of $G$ such that fixes every subgroup of
$G$, then what can we say about $g$ ?
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2 Answers
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It is called a power automorphism of the group $G$. The automorphism $g$ maps every element $x$ of $G$ to a power of $x$. See the following reference as a starting point.
Christopher D. H. Cooper, Power automorphisms of a group, Mathematische Zeitschrift, 1968, Volume 107, Issue 5, pp 335-356.
answered Nov 15, 2013 at 8:28
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In the case where $G$ is a $p$-group, we can say that the order of the automorphism $g$ has the form $p^er$, where $e \ge 0$ and $r$ divides $p-1$. Actually, more is true. Suppose we look at the group $A$ of $\,all\,$ automorphisms of $G$ that stabilize every subgroup of $G$. In the case that $G$ is a $p$-group, $A$ has a normal $p$-subgroup $N$ such that $A/N$ is cyclic of order dividing $p-1$. I won't go into the details of the proof here, but to get started, let me say that $N$ is the kernel of the action of $A$ on the Frattini factor group $P/\Phi(P)$. Then $A/N$ acts faithfully on this factor group and stabilizes all subgroups. Now think "linear algebra".
