Suppose I have a set $U$ in $\mathbb{R}^n$. For example, $[3,5]$ in $\mathbb{R}$. We know that the volume (length) in this example is $2$.
Suppose I have a permutation matrix $P$, I want to show the volume: $\mbox{vol}(U) = \mbox{vol}(PU)$. $PU $ means permuting each point in $U$. Here the permutation means $$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \rightarrow \begin{bmatrix} x_2 \\ x_1 \\ x_3 \end{bmatrix}. $$
It seems trivial because $|P| = \pm1$, it is an isometry and entries of $P$ are $1$ and $0$. So permutation matrices preserve volume.
However, is there any way to strictly prove it?