The problem
Let $\alpha$ (known), $\beta$ be permutations in $S_9$. Find $\beta$ such that $\alpha = \beta\alpha\beta^{-1}$
$\alpha = (178)(2593)(46)$
$\alpha^{-1} = (871)(3952)(46)$
The proposed solution
$\beta\alpha\beta^{-1} = (\beta(1)\beta(7)\beta(8))(\beta(2)\beta(5)...)(...)$. This becomes $\alpha^{-1}$ if you let $\beta(1)=1 \beta(7)=8 \beta(8)=7 \beta(2)=2 \beta(5)=5 ...$
This is an exam problem that I've tried to solve myself (and failed) and I can't follow the solution given, so I need some extra help in making sense of the method used.
The key point is the so-called conjugation formula:
Proof. To prove this identity, let $i\in\{1,\cdots,n\}$ and distinguish the following cases:
If $i$ is equal to $\sigma(a_j)$ for one $j\in\{1,\ldots,k\}$, then $(a_1,\cdots,a_k)$ sends $\sigma^{-1}(i)$ on $\left\{\begin{array}{cc}a_{j+1}&\textrm{if }j<k\\a_1&\textrm{if }j=k\end{array}\right.$. Therefore, one has: $$\sigma(a_1,\ldots,a_k)\sigma^{-1}=\left\{\begin{array}{cc}\sigma(a_{j+1})&\textrm{if }j<k\\\sigma(a_1)&\textrm{if }j=k\end{array}\right..$$
If $i$ is distinct from all the $\sigma(a_j)$, then $\sigma^{-1}(i)$ is fixed by $(a_1,\ldots,a_n)$ and one has: $$\sigma(a_1,\ldots,a_k)\sigma^{-1}(i)=\sigma(\sigma^{-1}(i))=i.$$ Notice that since $i$ is distinct from all the $\sigma(a_j)$, then $i$ is fixed by $(\sigma(a_1),\ldots,\sigma(a_k))$.
Whence the result. $\Box$
To conclude your exercise, one has to show that: $$\forall(\sigma,\tau_1,\tau_2)\in{S_n}^3,\sigma(\tau_1\tau_2)\sigma^{-1}=(\sigma\tau_1\sigma^{-1})(\sigma\tau_2\sigma^{-1}).$$ Which should be clear.
This approach is systematic to state whether or not two permutations are conjugated in $S_n$: