<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] Note that $S_6$ is the symmetric group of $6$ symbols which consists of $6!$ element.
We can use the theorem that two permutations are conjugate iff they have the same cycle type.
As you have already simplified the permutations into the product of disjoint cycles, it is easy to see that the two permutations have the same cycle type. Fortunately, using the theorem, no further calculations are needed.
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