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Let $G_1$ and $G_2$ be two isomorphic finite groups. Suppose that $\sigma_i:G_i\to G_i$ and $\sigma_i':G_i\to G_i$ be two isomorphisms for $i=1,2$.

When is there an isomorphism $\phi:G_1\to G_2$ such that $\phi\sigma_1=\sigma_2\phi$ and $\phi\sigma_1'=\sigma_2'\phi$?

Thanks for your helps in advance

P.S.

The origin of the problem:

Let $G=G_1\times G_2$ and suppose that $Aut(G)$ denote the automorphism group of $G$. Clearly every $\sigma\in Aut(G)$ such that $\sigma(G_1)=G_1$ and $\sigma(G_2)=G_2$ can induce a group automorphism of $G_1$ and $G_2$.

I try find a bijection $\phi$ from $G_1$ to $G_2$ such that $$\phi(\sigma(g_1))=\sigma(\phi(g_1))$$ for a subgroup $H$ of $Aut(G)$ such that every $\sigma\in H$ has the above property.

If for every $g_1\in G_1$ we have

$$ \phi(\sigma(\phi^{-1}(g_2)))=\sigma(g_2),$$ then it is a bijection with the above property.

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    $\begingroup$ Take $\sigma_1 = \sigma_1'$ and $\sigma_2, \sigma_2'$ distinct. $\endgroup$ – TastyRomeo May 22 '18 at 7:17
  • $\begingroup$ @TastyRomeo Thank you very much for your help but I mean for arbitrary $\sigma_i$ and $\sigma'_i$. $\endgroup$ – khers May 22 '18 at 7:45
  • $\begingroup$ @TastyRomeo Thanks so much. You mean that it is not always possible. I correct the question. $\endgroup$ – khers May 22 '18 at 7:51
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    $\begingroup$ You'll get some feeling for how complicated this is by looking at examples. For instance, try $G_1=G_2=\text{GL}(V)$ with $V$ a finite dimensional vector space. Answering the question whether there exists $\phi$ with $\phi^{-1}\sigma_1\phi=\sigma'_1$ is equivalent to the Rational Canonical Form Theorem -- non-trivial but within anyone's grasp. Now asking when we can do this simultaneously for a pair of $\sigma$'s: that's going to need a bit of work. $\endgroup$ – ancientmathematician May 22 '18 at 8:12
  • $\begingroup$ ps You said finite groups so $V$ had better be over a finite field. $\endgroup$ – ancientmathematician May 22 '18 at 8:22

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