Non-trivial automorphisms that fix maximal subgroups?

I thought this question up myself. It's somewhat similar to a question I asked previously Progress on a conjecture of Burnside...

For the relevant definitions consult basic texts on group theory. (I like Fraleigh, but there are plenty of others.)

Maximal subgroups are invariant under automorphisms. (This can be used to see that the Frattini subgroup is characteristic. And in fact that's what got me to thinking about it.)

Now, is every automorphism fixing the maximal subgroups the identity? By fixing I mean: maps each maximal subgroup to itself.

Checking in $\mathscr S_3$, for example, all the automorphisms are inner. So we have $6$ automorphisms to check.

There's work left to do. It's just a question of how much. I doubt this will be as difficult as the classification of finite simple groups, or Fermat's last theorem. But I don't actually know.

Any ideas?

This is of course naive of me. I haven't checked around that much. $\mathscr S_4$ is already more difficult, with $24$ automorphisms and maximal subgroups of orders $6,8$ and $12$.

So we can continue to look around for a counterexample; or try to prove a general result.

The question has been answered in the negative with @reuns's help. Cyclic groups provide a counter example, since there is only one subgroup of each order.

Don't know why I chose symmetric groups first. I appear to have overshot. No reason why an answer couldn't come from abelian groups (as it turns out to have).

Solution 1:

Let $(A, +)$ be an abelian group which contains maximal subgroups*. Consider the inversion map $x\mapsto -x$.

As $A$ is abelian, this is an automorphism. This map fixes all subgroups, not just the maximal ones. It is non-trivial if $A$ contains a non-trivial element of order other than $2$ (e.g. $A$ cyclic of order $\geq3$ works). Therefore, it satisfies all the properties you wish.

*e.g. not $\mathbb{Q}$, but any finitely generated abelian group works.