<s>[INST] Permutations conjugated 
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
  and
  $\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such that $\gamma\alpha\gamma^{-1}=\beta$

I know that $\alpha=(1,2,5)(4,6)(3)$ and $\beta=(2,3,4)(1,5)(6)$ also if $\sigma_1=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 5 & 4 & 6 \end{smallmatrix} \bigr)$ then $\sigma_1(1,2,5)\sigma_1^{-1}=(2,3,4)$; if $\sigma_2=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \end{smallmatrix} \bigr)$ then $\sigma_2(4,6)\sigma_2^{-1}=(1,5)$ and if $\sigma_3=(3,6)$ then $\sigma_3(3)\sigma_3^{-1}=(6)$.
But I don't know how to build $\gamma$ such that $\gamma\alpha\gamma^{-1}=\beta$. Can you help me, please?
 [/INST] One solution is to rewrite $\gamma\alpha\gamma^{-1}=\beta$ as $\gamma\alpha=\beta\gamma$.
Then $\gamma\alpha(1)=\beta\gamma(1)$, and from $\alpha(1)=2$ we get $\gamma(2)=\beta\gamma(1)$.
Similarly, $\gamma(5)=\beta\gamma(2)$ and $\gamma(1)=\beta\gamma(5)$.
Combining these, we get $\gamma(2)=\beta\gamma(1)=\beta\beta\beta\gamma(2)$, so $\gamma(2)$ has order 3 under $\beta$, and the same is true for $\gamma(5)$ and $\gamma(1)$.
Can you take it from here?
</s>