I am given two permutations
$$\alpha = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 5 & 6 & 7 & 8 & 3 & 11 & 9 & 12 & 1 & 10 & 2 & 4\end{pmatrix} $$ and $$\beta = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 6 & 7 & 4 & 5 & 12 & 1 & 11 & 9 & 2 & 3 & 10 & 8\end{pmatrix}. $$ Now I need to find a cyclic structure of permutation $\pi$, for which holds $$\beta^{-1}\pi^{2016}\beta = \alpha$$ and then find two distinct solutions.
My approach was to first write the $\alpha$ and $\beta$ as cycles. Then I calculated $\pi^{2016} = \beta \alpha \beta^{-1}$ and I got $$\pi^{2016} = \begin{pmatrix} 1 &10&7\end{pmatrix}\begin{pmatrix}2&6&12&4&11\end{pmatrix}\begin{pmatrix}5&9&8\end{pmatrix}.$$
Since $\pi^{2016}$ has a $5$-cycle and $2016 = 2015 + 1$ and $5 \mid 2015$, $\pi$ has the same $5$-cycle. But because $2016$ is divisible by $3$ and $6$, $\pi$ cannot have a $3$ or $6$-cycle. Using a theorem that says that cycle of length $L = mn$ raised to $m$-th power, decomposes into $m$ cycles of length $n$, no other $L$'s, $L < 7$ can decompose into two $3$-cycles. I am kinda stuck here...
The problem is that as a math major we did not do any sorts of such things yet, I assume because I did not take number theory. And in algebra we were just doing proof over proof and no concrete task.
I assume this task is a very easy one but please understand my situation.