The problem is this: prove that the two permutations, $x=(1\ 2\ 5)(6\ 4)$ and $y=(2\ 3\ 4)(1\ 5) $ are conjugates in the group $S_6$.
I was trying to find a element $z$ such that $zxz^{-1} =y$, but am unsure how to start. How are you supposed to find $z$? Is there some kind of formula or systematic method?
A followup question was to find how many $ z$ exist such that $zxz^{-1} =y$.
Hint: For all $\sigma,\tau\in S_n$ for any $n$, with
$$\tau=(t_{i_1}t_{i_1+1}\cdots t_{i_1+k_1})\cdots (t_{i_j}t_{i_j+1}\cdots t_{i_j+k_j})$$
we have
$$\sigma\tau\sigma^{-1}=(\sigma(t_{i_1})\sigma(t_{i_1+1})\cdots\sigma (t_{i_1+k_1}))\cdots (\sigma(t_{i_j})\sigma(t_{i_j+1})\cdots\sigma( t_{i_j+k_j})),$$
where $\sigma(x)$ is $\sigma$ evaluate at $x$.
For example,
$$\begin{align} ((123)(45))\circ (\color{red}{12})\circ((123)(45))^{-1}&=(((123)(45))(\color{red}1)((123)(45))(\color{red}2))\\ &=(23), \end{align}$$
which checks out, since
$$\begin{align} 1&\xrightarrow{((123)(45))^{-1}}3\xrightarrow{(12)}3\xrightarrow{(123)(45)}1\\ 2&\xrightarrow{((123)(45))^{-1}}1\xrightarrow{(12)}2\xrightarrow{(123)(45)}3\\ 3&\xrightarrow{((123)(45))^{-1}}2\xrightarrow{(12)}1\xrightarrow{(123)(45)}2\\ 4&\xrightarrow{((123)(45))^{-1}}5\xrightarrow{(12)}5\xrightarrow{(123)(45)}4\\ 5&\xrightarrow{((123)(45))^{-1}}4\xrightarrow{(12)}4\xrightarrow{(123)(45)}5 \end{align}$$