I am trying to understand in standard terms the example that is used to explain conjugation of permutations in here. Since the explanation is trying to avoid any prior knowledge of conjugation, it is hard to reproduce in a short, concise way, but it is based on shuffling cards that are indexed by the numbers in the gray rows (below), followed by relabeling. The steps are summarized in the presentation as follows:
The stacked-up notation (presumably) would be
$$\begin{align} \sigma &= \begin{pmatrix} 0\; 1 \; 2 \; 3 \; 4 \\ 2 \; 4 \; 1 \; 0 \; 3 \end{pmatrix}\\\\ \tau\sigma\tau^{-1} &= \begin{pmatrix} 0\; 1 \; 2 \; 3 \; 4 \\ 1 \; 3 \; 4 \; 2 \; 0 \end{pmatrix}\\ \end{align}$$
with $\sigma$ being the original permutation, and $\tau$ the permutation that outputs the conjugation.
In cycle notation
$$\begin{align} \sigma &=(0 \; 2 \; 1 \; 4 \; 3)\\ \tau\sigma\tau^{-1} &=(0 \; 1 \; 3 \; 2 \; 4) \end{align}$$
If this makes any sense, what would be the permutation $\tau$ used in the conjugation?
As per the comment, Tau can be obtained by stacking up
$$\begin{bmatrix}\begin{align}\sigma= (0\,2\,1\,4\,3)\\\tau\sigma\tau^{-1}=(0\,1\,3\,2\,4) \end{align}\end{bmatrix}\implies \tau =(2\,1\,3\,4) $$
For this to be concordant with the card game on the presentation the first row (first permutation given) should be understood as $0$ moves to slot number $3$; $1$ moves to slot number $2$; $\quad 2 \to 0; \quad 3 \to 4; \quad 4 \to 1$, which would result in
$$\sigma=(0\,3\,4\,1\,2)$$
And, similarly,
$$\tau\sigma\tau^{-1}=(0\,4\,2\,3\,1)$$
which makes it all fall in place when stacking them up to get tau:
$$\tau=(0)\,(3\,4\,2\,1).$$