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?

Let $\gamma$ be the function that takes a number in the cycle of $\beta$ to the number in $\alpha$ in the corresponding position. That is, send $2 \rightarrow 1$, and $3 \rightarrow 2$, and $4 \rightarrow 5$ et cetera. 
Intuitively you can think of this as as you temporarily relabelling the numbers in $\beta$ (applying $\gamma$), applying the permutation $\alpha$ and then "undoing" this relabelling at the end ( applying $\gamma ^{-1} $).
Of course, the choice of $\gamma$ isn't unique. We could have sent $2 \rightarrow 2$, $3 \rightarrow 5$ and $4 \rightarrow 1$, for example. 
This method should work in general for finding $\gamma$ for any permutations of any order and type.
