Is there a way to determine if an even permutation $a$ exists such that $a\tau a^{-1}=\sigma$

Question

I was looking at an old algebra exam, and one of the problems were to determine if there exists a permutation $$a\in A_{10}: a\tau a^{-1}=\sigma$$ where $\sigma=(1\space7)(5\space6\space9)(\space2\space3\space4\space8\space 10), \tau=(1\space2)(3\space4\space5)(6\space7\space8\space9\space10)$. And I'm not really sure where to start. I have tried taking the sgn function on both sides and moving things around, but this doesn't seem to get me anywhere. Is there a general way of determining wether such an $a$ exists?

Answer 1

Yes, such an $a$ must exist. Since $\sigma$ and $\tau$ have the same cycle structure, $\exists b \in S_{10}~(b \tau b^{-1} = \sigma)$. If $b$ is even, you're done. If $b$ is odd, then $a=b \circ (1~2)$ is even and $(1~2)$ commutes with $\tau$. This argument works as long as $\tau$ contains at least one cycle of even length. A very similar argument also works if $\tau$ contains at least two cycles of the same odd length.

Answer 2

Here is one such even permutation,

$$\sigma =(126489537)$$

You line up the cycles structures, and then it is a question of whether to use $(12)$ or $(21)$, the latter gives an odd cycle.