I am given two permutations in $S_{10}$: $$\alpha = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 & 4 & 8 & 6 & 7 & 10 & 1 & 9 & 3 & 2 \end{pmatrix}$$
$$\beta = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 3 & 7 & 5 & 6 & 9 & 8 & 2 & 10 & 1 \end{pmatrix}$$
and my question is: how do I find $k>0$ such that $\alpha ^k$ and $\beta ^k$ conjugate and are not the identity permutation? I tried doing it and I know that the cycles should be the same, but I find that $k=2$ gives us the same cycles and $k=4$ also, and I am not sure if I am correct or totally wrong.
I know that two permutations $\sigma,\sigma′\in S_n$ are conjugate if exists $\tau\in S_n$ such that: $\sigma′=\tau\sigma\tau−1=(\tau(a_0),\tau(a_1)…\tau(a_k))$, where $\alpha=(a_0a_1…a_k)$.
Is there an easy way to find $k$ and $\tau$ that works in this case?
The two cycle structures are $(1\ 5\ 7)(2\ 4\ 6\ 10)(3\ 8\ 9)$ and $(1\ 4\ 5\ 6\ 9\ 10)(2\ 3\ 7\ 8)$.
Cubing gives the cycles $(2\ 10\ 6\ 4)$ and $(1\ 6)(4\ 9)(5\ 10)(2\ 8\ 7\ 3)$ and it is now obvious that a further squaring makes both permutations into two involutions and so $k=6$ works.
However, as you point out $k=2$ and $k=4$ also work. So perhaps the best answer is that $k$ can be any multiple of $2$ which is not also a multiple of $12$.