We have $\sigma = (13624)(587)(9)$ and $\tau=(15862)(394)(7)$. Determine a permutation $\alpha$ such that $\alpha\sigma\alpha^{-1} = \tau$. How many such $\alpha$ are there?
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The key to this question is to use the following general fact (see here for example):
If $\alpha, \sigma \in S_n$ and $\sigma = (x_{11}\ \dots\ x_{1k_1})\dots(x_{l1}\ \dots\ x_{lk_l})$ is its cycle decomposition, then $$\alpha\sigma\alpha^{-1} = (\alpha(x_{11})\ \dots\ \alpha(x_{1k_1}))\dots(\alpha(x_{l1})\ \dots\ \alpha(x_{lk_l})).$$
With this fact at your disposal, the equation $\alpha\sigma\alpha^{-1} = \tau$ becomes
$$(\alpha(1)\ \alpha(3)\ \alpha(6)\ \alpha(2)\ \alpha(4))(\alpha(5)\ \alpha(8)\ \alpha(7))(\alpha(9)) = (1\ 5\ 8\ 6\ 2)(3\ 9\ 4)(7).$$
By matching elements of the cycles, we see that $\alpha$ given by
$$\begin{array}{ccccccccc} 1&2&3&4&5&6&7&8&9\\ 1&6&5&2&3&8&4&9&7 \end{array}$$
is such a permutation. In cycle notation, we have $\alpha = (2\ 6\ 8\ 9\ 7\ 4)(3\ 5)(1)$. Note, this isn't the only possible solution. For example, $(1\ 2\ 8\ 9\ 7\ 4\ 6\ 5\ 3)$ also works.
So the question is now, how many choices of $\alpha$ are there? Suppose $\beta\sigma\beta^{-1} = \tau$, then $\alpha\sigma\alpha^{-1} = \beta\sigma\beta^{-1}$. Rearranging this equation, we get $\alpha^{-1}\beta\sigma\beta^{-1}\alpha = \sigma$ so $\alpha^{-1}\beta \in C_{S_n}(\sigma)$, the centraliser of $\sigma$ in $S_n$, and therefore $\beta \in \alpha C_{S_n}(\sigma)$. So we have a bijection
$$\{\beta \in S_n \mid \beta\sigma\beta^{-1} = \tau\} \to \alpha C_{S_n}(\sigma)$$
and therefore the number of choices of $\alpha$ is equal to $|\alpha C_{S_n}(\sigma)| = |C_{S_n}(\sigma)|$. By the Orbit-Stabiliser Theorem,
$$|C_{S_n}(\sigma)| = \frac{|S_n|}{|\operatorname{Orb}(\sigma)|} = \frac{n!}{|[\sigma]|}$$
where $[\sigma]$ denotes the conjugacy class of $\sigma$ in $S_n$. As two elements of $S_n$ are conjugate if and only if they have the same cycle type, we need to calculate the number of permutations which have the same cycle type as $\sigma$. So far what has been applied works in general, but let us now consider the case of the specific $\sigma$ in the question. Note that $\sigma$ has cycle type $(5, 3, 1)$; how many permuations hace the same cycle type? This is equivalent to asking how many ways there are of dividing $9$ objects into a group of $5$, a group of $3$ and a group of $1$. The answer is given by a multinomial coefficient, namely
$$\displaystyle\binom{9}{5,\ 3,\ 1} = \dfrac{9!}{5!\,3!\,1!}.$$
Using this we have
$$|\{\beta \in S_9 \mid \beta\sigma\beta^{-1} = \tau\}| = |C_{S_9}(\sigma)| = \frac{9!}{|[\sigma]|} = \frac{9!}{\displaystyle\binom{9}{5,\ 3,\ 1}} = \frac{9!}{\dfrac{9!}{5!\,3!\,1!}} = 5!\,3!\,1! = 720.$$
Michael Albanese
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