Approaching Symmetric Groups-questions So I'm trying to get into Symmetric Groups in my Discrete Mathematics course and I have a problem that I need some help on. 
I have a permutation $\alpha = (178)(2593)(46)$ and its inverse $\alpha^{-1} = (187)(2395)(46)$. I'm supposed to find a function $\beta$ so that $\alpha^{-1} = \beta*\alpha*\beta^{-1}$.
I have the solution (or one possible, anyway) and its $\beta = (35)(78)$ but I want to understand how they came to that conclusion. 
What's the go-to-algorithm/approach when faced with similar problems?
 A: The thing to notice is that given some permutations $\sigma=(a_1,a_2,\ldots,a_n)$ and an arbitrary $\tau$ and element $i$, 
\begin{align*}
(\tau\sigma\tau^{-1})(\tau i)&=(\tau(a_1,a_2,\ldots,a_n)\tau^{-1})(\tau i)\\
&=\tau(a_1,a_2,\ldots,a_n)(i)\\
&=(\tau a_1,\tau a_2,\ldots,\tau a_n)(i)
\end{align*}
so conjugating by $\tau$ is just a renaming of elements. For example,
$$(1,4)(1,2,3,4,5)(1,4)=(4,2,3,1,5).$$
So now it should be pretty simple, we have $\alpha=(1,7,8)(2,5,9,3)(4,6)$ and $\alpha^{-1}=(1,8,7)(2,3,9,5)(4,6)$, so by the above we know that 
$$\tau=(1)(7,8)(2)(5,3)(9)=(7,8)(5,3)$$
will do the trick.
A: Hope this helps visualizing: you should think about conjugation of permutation $\beta \alpha \beta^{-1}$ as "renaming" the numbers in $\alpha$ according to $\beta$. In your example: Take the permutation $(178)(2593)(46)$ and switch $3$ with $5$ and $7$ with $8$. The result is $(187)(2395)(46)$. That means conjugation always leaves the "structure" of the permutation the same (i.e. the cycle-lengths), but only changes which numbers are which.
