# Example of non-preservation of isomorphism classes of $G$-sets under pullback along automorphisms of $G$, with $G$ simple?

Definition: if $$G$$ is a group, $$X$$ is a left $$G$$-set, and $$\rho$$ is a group automorphism of $$G$$, let us say the data $$(G,X,\rho)$$ has property $$P$$ if $$\rho^* X$$ is not isomorphic to $$X$$ as a left $$G$$-set. (Here $$\rho^*X$$ is the left $$G$$-set with underlying set the same as for $$X$$, and with $$g$$ acting on $$\rho^*X$$ as $$\rho(g)$$ acts on $$X$$.)

I'm looking for examples of $$(G,X,\rho)$$ having property $$P$$; so far, the only example I've come up with is the following: if $$H$$ is some nontrivial group, if $$Y$$ is some set with free left $$H$$-action, if $$G := H \times H$$ as a direct product group with projection homomorphisms $$H \xleftarrow{\pi_1} G \xrightarrow{\pi_2} H$$ onto the 1st and 2nd components respectively, if $$\rho$$ is the automorphism of $$G$$ that exchanges the two $$H$$ factors (mapping each "identically" onto the other), if $$X := \pi_1^* Y$$, then $$\rho^*X \simeq \pi_2^* Y$$ and furthermore there is no $$G$$-equivariant map between $$X$$ and $$\rho^*X$$ (in either direction); in particular $$X$$ and $$\rho^*X$$ cannot be isomorphic as $$G$$-sets.

Would anyone know how to find an example of $$(G,X,\rho)$$ having property $$P$$ where $$G$$ is a simple group? (Or, how to show such an example doesn't exist?)

If $$\rho$$ is an automorphism and $$H$$ is a subgroup of $$G$$, then $$\rho^* (G/H) \cong G / \rho^{-1}(H)$$. This is isomorphic to $$G/H$$ iff $$H$$ and $$\rho^{-1}(H)$$ are conjugate. This yields many examples where they are not isomorphic. One case is when $$G$$ is abelian and $$\rho(H) \neq H$$.
• Thank you! However, I'm looking for an example where $G$ is simple (I've now updated the question title to make this more explicit). Would you have a suggestion on how to find an example of a simple group $G$, a group automorphism $\rho$ of $G$, and a subgroup $H \leq G$, such that $H$ and $\rho(H)$ are not conjugate in $G$? Commented Sep 4 at 22:56
• @I.A.S.Tambe: look at copies of $A_5$ in $A_6$ and use an outer automorphism of $S_6$. Commented Sep 4 at 23:21
• It's not a non-characteristic subgroup I don't think. You need at least a subgroup that is moved by an outer automorphism. But that's not enough either: for example, every two subgroups of $A_5$ that are conjugate via an outer automorphism, are also conjugate via an inner automorphism. Commented Sep 6 at 3:46