Mathematics Asked by Benjamin on October 3, 2020

Let $G$ be a finite non-cyclic group. Can a non-inner automorphism map every subgroup to its conjugate? Namely, can there be a non-inner automorphism $alpha$ that, for every $Hle G$, there exists some $g$ in $G$ such that $alpha(H)=H^g$?

Yes, the dihedral group of order 10 has this property. Its subgroup structure is very simple: $D_{10}$, ${e}$, the rotation subgroup, and the 5 subgroups generated by a flip. Any automorphism fixes the first three, so those are done, and shuffles the last five, and those 5 subgroups are all conjugate to each other.

All we need now is to show that there is a non-inner automorphism, but this is easy; the inner automorphisms always send a rotation to its inverse (or fix it) so we only need an automorphism which doesn't do that. Let the generators be $sigma,tau$, rotation and flip, and consider the automorphism defined on the rotations by $sigma mapsto sigma^2$ and fixing $tau$.

Correct answer by TokenToucan on October 3, 2020

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