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!
 A: There are more solutions than just $\tau=(17)(26)(35)$. Your task was only to find one such $\tau$, so that's fine, but it is not too hard to find them all. And in the process, see that the cycle structure of $\tau$ is unrelated to the cycle structure of your permutations.
The cycles in your given permutations are all disjoint, and that is important. $\tau$ must map the three elements of the triple $(123)$ to $7$, $6$, and $5$, but not necessarily in that order, since $(765)=(657)=(576)$. $\tau(1)$ could be $5$, $6$, or $7$. Based on the choice made, $\tau(2)$ and $\tau(3)$ are determined.
The other two cycles are both 2-cycles, and they are disjoint so they permute. So $\tau$ could be anything that takes $4$ and $5$ to either $4$ and $3$ or $2$ and $1$, and further there is not need to respect the internal order of these pairings ($(43)=(34)$).
Considering the three-cycle, we have one of the following: 
$$\begin{align}
\tau:&1\mapsto7&\tau:&1\mapsto6&\tau:&1\mapsto5\\
\tau:&2\mapsto6&\tau:&2\mapsto5&\tau:&2\mapsto7\\
\tau:&3\mapsto5&\tau:&3\mapsto7&\tau:&3\mapsto6\\
\end{align}$$
Considering the two-cylces, we have one of the following:
$$\begin{align}
\tau:&4\mapsto4&\tau:&4\mapsto4&\tau:&4\mapsto3&\tau:&4\mapsto3\\
\tau:&5\mapsto3&\tau:&5\mapsto3&\tau:&5\mapsto4&\tau:&5\mapsto4\\
\tau:&6\mapsto2&\tau:&6\mapsto1&\tau:&6\mapsto2&\tau:&6\mapsto1\\
\tau:&7\mapsto1&\tau:&7\mapsto2&\tau:&7\mapsto1&\tau:&7\mapsto2\\
\end{align}$$
This gives $12$ possible maps for $\tau$. Calculating each one's cycle structure, the possibilities for $\tau$ are: $$(17)(26)(35), (1726)(35), (17)(26)(354), (1726)(354), (162537), (16)(2537), (1625437), (16)(25437), (153627), (1536)(27), (1543627), (15436)(27)$$
A: Conjugation permutes elements "in place" in their cycle structure.  Hence you want $\tau$ to map 1 to 7, 2 to 6, 3 to 5, 4 to 4, 5 to 3, 6 to 2, and 7 to 1.  That is, $$\tau=(17)(26)(35)$$
A: 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.
