Does there exist an explicit criterion (or a good sufficient condition) for proving that a Vandemonde matrix: $$A(x_1,\dots,x_n):=\left[ \begin{array}{llll}1 & x_1 &\dots& x_1^{n-1}\\ 1 & x_2 &\dots& x_2^{n-1}\\&&\vdots&\\ 1& x_n &\dots& x_n^{n-1}\end{array}\right]$$ has nonzero permanent? ($x_1,\dots,x_n$ are complex numbers) I'm especially interested in the case where $x_i$'s are roots of unity.

Added. Here is an explicit conjecture I want to prove:

If $x_1,\dots,x_n$ are in $\mu_N$ (not necessarily different) and $n$ is coprime to $N$ then $A(x_1,\dots,x_n)\neq 0$.

The conjecture is trivially true when $N$ is a prime and $n<N$, since $A(x_1,\dots,x_n)$ is a sum of $n!$, $N$th roots of unity and $N$ dose not divide $n!$.