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?
 A: 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.
A: 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.
A: 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?
A: $\;\gamma^{-1}\;$ will be the permutation mapping each cycle of $\;\alpha\;$ to a corresponding cycle of $\;\beta\;$ of the same type, thus for example:
$$\gamma^{-1}:=\begin{cases}1\to2\\2\to3\\5\to4\\.......\\4\to1\\6\to5\\.......\\3\to6\end{cases}\implies\gamma=(123654)$$
and we can now check that
$$\gamma^{-1}\alpha\gamma=(123654)(125)(46)(3)(145632)=(15)(234)(5)=\beta$$
