How to solve equation $\tau(123)(45)(67)\tau^{-1}=(765)(43)(21)$ and alike? Let $\sigma\in S_7$ be $(123)(45)(67)$. And Find $\tau\in S_7$ such that $\tau\sigma\tau^{-1}=(765)(43)(21)$. I understand that in symmetric group conjugate elements have the same cycle structure. Hence, $\tau$ should share the structure of $\sigma$. Then I guess I could trial and error to find the solution for $\tau$. However, it is not very nice to do so. I am wondering whether there is a good way to solve such equation in general case. Thank you!

I am assuming that you multiply permutations from right to left like function composition.
Let $\sigma,\tau\in S_{n}$, and Let $\sigma(i)=j$.  Then $\tau\sigma\tau^{-1}\tau(i)=\tau(j)$.  Therefore $\tau$ acts on the "display" of $\sigma$.  So and say $\sigma=(a_{1}\cdots a_{r})$ then $\tau\sigma\tau^{-1}=(\tau(a_{1})\cdots\tau(a_{r}))$
For example, if $\sigma=(1 \ 2)(3 \ 4)$ and $\tau=(2 \ 4)$ then $\tau \sigma \tau^{-1}= (\tau(1) \ \tau(2))(\tau(3) \ \tau(4))=(1 \ 4)(3 \ 2)$.
In your case $\tau(1) =7$.  I will leave the rest to you.
