# Permutation isomorphic subgroups of $S_n$ are conjugate

Consider $G,H \leq S_n$ and their natural action on $[n] = \{1,\ldots,n\}.$ We say that $G$ and $H$ are permutation isomorphic if there is a bijection $\varphi:[n] \mapsto [n]$ and group isomorphism $f:G \mapsto H$ so that $$\varphi(g(o)) = f(g) (\varphi(o))$$ or in the standard notation involving group actions $\varphi(o^g) = \varphi(o)^{f(g)}.$

I would like to show that in this case $G$ and $H$ are in fact conjugate subgroups of $S_n.$ And that the converse is true as well.

I do not know how to really think about these problems. As I understand isomorphism of $G$ and $H$ does not imply conjugacy in $S_n$ but I have no idea how to use the relabelling property to force conjugacy?

Any ideas how to prove this?

Note that $\phi$ is itself an element of $S_n$, and that composition is the group operation. Then your equation becomes $\phi g=f(g)\phi$. If you need more hints, let me know.

• That does it, thanks. – Jernej Feb 5 '15 at 16:13