Solving for a Permutation? The question I'm asked to solve is as follows.

Give an example of the following with a brief justification:  A permutation $\pi \in S_4$ such that $(243) =\pi(123)\pi^{-1}$$


So far, I know that I can write $\pi$ as
$$\pi = \left[\begin{array}{c c c c} 1 & 2 & 3 & 4 \\ a & b & c & d\end{array}\right]$$
and I can rewrite the above equation as
$$(2,4,3)=\pi(1,2,3)\pi^{-1} \rightarrow (2,4,3)\pi = \pi(1,2,3)$$
From there I think I can write the left side of the equation as
$$(2,4,3)\pi = (2,4,3)\left[\begin{array}{c c c c} 1 & 2 & 3 & 4 \\ a & b & c & d\end{array}\right] = \left[\begin{array}{c c c c} 1 & 2 & 3 & 4 \\ a & d & b & c\end{array}\right]$$
but from here I'm unsure where to go. What should I do next?
 A: A very useful fact is that for any cycle $(a_1a_2a_3\cdots)$, we have $$\pi(a_1a_2a_3\cdots)\pi^{-1}=(\pi(a_1)\pi(a_2)\pi(a_3)\cdots)$$
So, we need to solve $(243)=\pi(123)\pi^{-1}=(\pi(1)\pi(2)\pi(3))$. We can therefore choose $\pi=(124)$ as a solution by lining up the indices, i.e. saying $\pi$ sends $(\color{blue}{123})$ componentwise to $(\color{red}{243})$: so $\color{blue}1\mapsto \color{red}2$, $\color{blue}2 \mapsto \color{red}4$, $\color{blue}3 \mapsto \color{red}3$. In matrix form, this would be $\pi=\left[\begin{matrix}1 &2 &3 &4\\ 2 &4 &3 &(?)\end{matrix}\right]$, and there's exactly one way to fill in the $(?)$.
But this isn't the only solution! Cycle notation is overspecified, so e.g. $(243) = (432) = (324)$. We could have gotten the solutions $\pi = (14)(23)$, $\pi = (134)$ from lining up these other two representations with $(\pi(1)\pi(2)\pi(3))$ and following the same procedure.
Note also that if we were dealing with, say $S_5$, we wouldn't have been able to uniquely determine where $4$ should have gone—it could just as well have gone to $5$, or something else if we had more indices to play with!
