If $|\tau|=2$ and $\sigma\tau\sigma=\tau$ in $S_8$ what are the possible order of $\sigma$? Let us consider $S_8$ the symmetric group of all permutation of the symbols $\{1, 2, ..., 8\}$ of order $8!$. And given that for some $\tau\in S_8$ of order 2, there exists $\sigma\in S_8$ such that $\sigma\tau\sigma=\tau$ holds. 
What are the possible choice of $|\sigma|$?
Note that here $|\theta|$ denotes the order of the permutation $\theta\in S_n$.
My try: If $\sigma$ and $\tau$ be disjoint permutation meaning by $\tau$ and $\sigma$ no common symbols from $\{1, 2, ..., 8\}$ then they commute. So that we shall get $\sigma\tau\sigma=\tau$ i.e. $\tau\sigma^2=\tau$ i.e. $\sigma^2=\varepsilon$ which shows that $\sigma$ must be a permutation of order 1 or 2. We are done in this case.
But what if $\sigma$ and $\tau$ share some common entry/ symbol?
Then ? What to do ? 
 A: Since $\tau=\tau^{-1}$, therefore we can rewrite $\sigma\tau\sigma=\tau$ as $\tau \sigma \tau^{-1}=\sigma^{-1}$. Thus we want those $\sigma$'s which upon conjugation by $\tau$ give $\sigma^{-1}$. Note that in the disjoint cycle representation of $\sigma$ and $\sigma^{-1}$ not only the cycle structure remains the same but the symbols within each cycle will also be the same (except in the ordering).
Let $\tau=(ij)$.


*

*If $i,j$ are not in any of the cycles of the disjoint cycle representation of $\sigma$ then $\tau\sigma\tau^{-1}=\sigma$. In which case we will have $\sigma=\sigma^{-1}$. Thus $|\sigma|=1 \text{ or }2$.

*If only one of $i$ or $j$ is in the disjoint cycle representation of $\sigma$ then the resulting permutation $\tau\sigma\tau^{-1} \neq \sigma^{-1}$ because either $i$ will be replaced by $j$ or vice versa and the conjugate of $\sigma$ will no longer have the same symbols in the cycle structure.

*So both $i,j$ should be in the the disjoint cycle representation of $\sigma$.

*But if $i$ and $j$ were in different cycles of the the disjoint cycle representation of $\sigma$ then too upon conjugation the symbols in the cycle will change and it can no longer be equal to $\sigma^{-1}$.

*So both $i,j$ have to be in the same cycle of the disjoint cycle representation of $\sigma$. If that cycle has length $\geq 4$, then by the exchange of $i$ with $j$, it no longer will result in the cycle structure that can possibly give $\alpha^{-1}$. Therefore that cycle can only be of length $2$ or $3$. The case of length $2$ can be easily dealt with. So let us consider the cycle of length $3$. If it is of the form $( i k j)$, then after the action of $\tau$ via conjugation it will become $(jki)$ which is the inverse of $(ijk)$. Now all the other cycles of the disjoint cycle representation of $\sigma$ cannot be of length more than $2$ (because if they were then upon conjugation by $\tau$ they will remain unchanged and hence cannot give us $\sigma^{-1}$. Thus the order of such a $\sigma$ can be only either $3$ or $6$.


Thus $|\sigma|=1,2,3,6$.
A: Given $\tau^2=1$. Then $\sigma\tau\sigma=\tau$ is equivalent to $\sigma\tau\sigma\tau=1$ i.e. $(\sigma\tau)^2=1$. 
In dihedral groups of order $2n$, we have a situation where $r$ and $sr$ are both reflections, and then $s$ is a rotation, whose possible orders are $n$ or its divisors. 
Thus, we can think of which dihedral groups can be embedded in $S_8$. The answer is: the dihedral groups of order $16,14,...,6$ (and of course $4,2$) can be embedded in $S_8$. 
Then for taking $\sigma$ to be rotation of maximum order in the diedral groups (considered as subgroup of $S_8$) we can see that the orders of these $\sigma$ are 
$\frac{16}{2}=8, \frac{14}{2}=7,\cdots, 1$. 
Thus, $1,2,...,8$ are among possible orders of $\sigma$.
I think, these are the only possible orders: since, a finite group generated by two elements of order $2$ is dihedral. (see these links 
1, and 2).
Let me know if some part is not clear.
