There exist $\beta \in$ Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ such that $\phi'(f)=\beta \circ \phi(f) \circ \beta^{-1}$ Let $\phi, \phi' : \mathbb{Z}/3\mathbb{Z} \rightarrow$ Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ 2 non-trivial homomorphisms. Show that there exist $\beta \in$ Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ such that $\phi'(f)=\beta \circ \phi(f) \circ \beta^{-1}$ $\forall f \in \mathbb{Z}/3\mathbb{Z}$. We know that Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong GL(2,\mathbb{F}_2) \cong S_3$ and that we must have $\phi([1]_3) = \sigma \in S_3$ such that $\sigma$ is a 3-cycle (if not we have a trivial homomorphism). Then i have problem to show that there exist $\beta \in$ Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ such that $\phi'(f)=\beta \circ \phi(f) \circ \beta^{-1}$ $\forall f \in \mathbb{Z}/3\mathbb{Z}$. I don't really see what happens when we want to ''jump'' from $S_3$ to Aut$(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$. Thanks in advance
 A: It was proven in that question that $\mathrm{Aut}(\Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z)$ is isomorphic to $\mathcal S_3$. Therefore, it is enough to prove the result for $\mathcal S_3$ instead of $\mathrm{Aut}(\Bbb Z / 2 \Bbb Z \times \Bbb Z / 2 \Bbb Z)$.
A non-trivial homomorphism $\phi \in \mathrm{Hom}(\Bbb Z / 3 \Bbb Z, \mathcal S_3)$ is fully defined by $\phi(1)$ as $\Bbb Z / 3 \Bbb Z$ is cyclic. And $\phi(1)$ is a 3-cycle $\sigma_\phi$as the 3-cycles are the only elements of order 3 in $\mathcal S_3$. In $\mathcal S_3$, there are only two 3-cycles. Namely $\sigma_1 = (1 \ 2 \ 3)$ and $\sigma_2 = (1 \ 3 \ 2)$. Therefore $\mathrm{Hom}(\Bbb Z / 3 \Bbb Z, \mathcal S_3)$ also has only two non-trivial elements $\phi_1,\phi_2$ with $\phi_1(1) = \sigma_1$ and $\phi_2(1) = \sigma_2$.
Now in $\mathcal S_n$, any two cycles of the same lengths are conjugate. Applying the result to our specific case, it exists $\beta \in \mathcal S_3$ such that $\sigma_2 = \beta \circ \sigma_1 \circ \beta^{-1}$. By the way $\beta = (2 \ 3)$ is such an element.
This conjugacy relation means that
$$\phi_2(1) = \beta \circ \phi_1(1) \circ \beta^{-1}$$ and implies
$$\phi_2(2) = \phi_2(1+1)=(\beta \circ \phi_1(1) \circ \beta^{-1})^2=\beta \circ (\phi_1(1))^2 \circ \beta^{-1} = \beta \circ \phi_1(2) \circ \beta^{-1}$$
As $$\phi_2(0) = \mathrm{Id}= \beta \circ \phi_1(0) \circ \beta^{-1}$$
we have proven the desired result.
