Recall that a Vandermonde matrix is a matrix of the form
$$V(\vec x)=V(x_0,...,x_n):=\begin{pmatrix} 1 & x_0 & x_0^2 & \cdots & x_0^n \\ 1 & x_1 & x_1^2 & \cdots & x_1^n \\ \vdots &\vdots &\vdots & & \vdots &\\ 1 & x_n & x_n^2 & \cdots & x_n^n \\ \end{pmatrix}$$
for numbers $x_0,...,x_n\in\Bbb C$, and that $V(\vec x)$ is invertible if and only if the $x_i$ are distinct.
I am interested in the case when the inverse of a Vandermonde matrix is again a Vandermonde matrix, or $V(\vec x)V(\vec y)=\mathrm{Id}$. Comparing number of variables ($=2(n+1)$) with number of constraints ($=(n+1)^2$) we should not expect there to be solutions, except for $n\in\{0,1\}$.
Here is a solution for $n=1$: $\vec x=\vec y=(0,-1)$, or
$$V(\vec x)=V(\vec y)=\begin{pmatrix} 1 & \phantom+0 \\ 1 & -1 \end{pmatrix}.$$
Question: Can the inverse of a Vandermonde matrix be a Vandermonde matrix for $n\ge 2$?