It says here that $\rho = \tau \sigma \tau^{-1}$, and if $\sigma(i) = j$, then $\rho(\tau(i)) = \tau(j)$. So take for example $\sigma = (1 3 2 4) (5 6)$ and $\rho = (5 2 3 1)(6 4)$. To find $\tau$, it says that you can just line up $\sigma$ and $\rho$ like so $$\begin{array}{cccccc} 1&3&2&4&5&6\\ 5&2&3&1&6&4 \end{array}$$ and take $\tau$ to be $(1 5 6 4)(2 3)$. I don't understand why lining up the permutations like this works in finding $\tau$, as it just looks like magic. The fact that $\rho(\tau(i)) = \tau(j)$ and $\rho = \tau \sigma \tau^{-1}$ doesn't seem to explain this phenomenon to me, while textbooks/online resources just brush it off as obvious as a result of the aforementioned equations, so is there an intuitive explanation for why this works?
i.e. If $\sigma(a_1) = a_2$, then $\tau \sigma \tau^{-1} (\tau(a_1)) = \tau(a_2)$. Why does this imply that $\tau \sigma \tau^{-1} = (\tau(a_1) \tau(a_2) ...)$?
Let $\sigma,\tau$ be a permutations on $\{1,\ldots, n\}$. If $\sigma$ is a cycle, say $\sigma=(i_1,\ldots i_k)$, then
$$\begin{align} \tau\sigma\tau^{-1}(\tau(i_j)) & = \tau\sigma(i_j)\\ & = \tau(i_{j'})\\ \end{align}$$ where $$j'=\begin{cases}j+1, & 1\le j<k\\1, & j=k\end{cases}.$$
So, $$\tau\sigma\tau^{-1}=(\tau(i_1)\tau(i_2)\ldots\tau(i_k)).$$