We will prove the following stronger statement: For any positive integer $n$ and any two unimodal arrangements of $n$ pupils, there is a sequence of swaps leading from one of the arrangements to the other, via unimodal arrangements.

We show this by induction on $n$. The result plainly holds when $n = 1$, as the sequence is both increasing and decreasing so no swap is required.

Suppose now that $n \ge 2$ and inductively we may assume the claim for $n-1$ pupils. The shortest pupil must be at one of the two ends in a unimodal arrangement, so we can partition the set of all unimodal arrangements of $n$ pupils into two subsets: $\mathcal{L}$ with the shortest pupil at the left end, and $\mathcal{R}$ with the shortest pupil at the right end.

Inductively, I claim that $\mathcal{L}$ is connected by swaps. Indeed, any two arrangements in $\mathcal{L}$ can be connected by swaps involving the taller $n-1$ pupils, since any unimodal arrangement of the $n-1$ taller pupils remains unimodal when the shortest pupil is placed at the far left. By the same argument, $\mathcal{R}$ is connected.

To complete the induction, we need to show that $\mathcal{L}$ is connected to $\mathcal{R}$. It suffices to describe one arrangement from $\mathcal{L}$ and one from $\mathcal{R}$ which are connected by a swap. For example, numbering the pupils from $1$ (shortest) to $n$ (tallest), we obtain an arrangement in $\mathcal{L}$ when we arrange the odd numbers in increasing order from the left, and the even numbers in increasing order from the right. Swapping pupils $1$ and $2$ gives an arrangement in $\mathcal{R}$.

This completes the inductive step and the result is proved.