Hall not implies automorph-conjugate

Statement

A Hall subgroup of a group need not be conjugate to all its automorphs.

Proof

We prove that if $r$ is an odd prime, $q$ is a power of a prime $p$, and $gcd(r,q-1)=1$, then any subgroup of index $(q^r-1)/(q-1)$ in $SL(r,q)$ is a Hall subgroup.

This follows from order computation.

Now observe that the parabolic subgroup $P_{r-1,1}$ has the required index, and hence is a Hall subgroup. By $P_{r-1,1}$ we mean the subgroup of $SL(r,q)$ comprising those elements where the bottom row has only one nonzero entry, namely the last.

Now consider $P_{r-1,1}$ and its image under the transpose-inverse automorphism. For $r>2$ (which is true if $r$ is an odd prime, the transpose-inverse has an invariant one-dimensional subspace while the original subgroup doesn't. Hence, the two subgroups cannot be conjugate. However, they are certainly automorphs (by the transpose-inverse automorphism). We thus have a Hall subgroup that is not automorph-conjugate.

A specific example is $SL(3,2)$, where the two Hall subgroups are both isomorphic to $S_4$.