Vandermonde $V_n$ mod $n$ Consider the all-familiar Vandermonde determinant $V_n(x_1,\dots,x_n)$ of the matrix of $(i,j)$-entries $M_n(i,j)=x_j^{i-1}$ so that
$$V_n(x_1,\dots,x_n)=\prod_{1\leq i<j\leq n}(x_j-x_i).$$
Let's specialize the variables $x_k=k\pi_k$ where $\pi\in\mathfrak{S}_n$ is a permutation of $n$ letters $\{1,2,\dots,n\}$.
Assume $n>3$. I like to ask:

QUESTION. Is this true? $V_n(1\pi_1,2\pi_2,\dots,n\pi_n)$ is congruent to $0$ moduluo $n$ for any permutation $\pi\in\mathfrak{S}_n$.

 A: Per comments above, for a counterexample we have with necessity $\pi_n=n$ and prime $n$. The case $n=2$ is trivial, so I assume that $n$ is an odd prime.
The elements $U:=\{ 1,2,\dots,n-1\}$ form the unit group of $GF(n)$ and the mapping $i\mapsto i\pi_i$ has to be a permutation of $U$. Such mappings are called complete and it's known that they do not exists whenever the group has a nontrivial, cyclic Sylow 2-subgroup. In our case, $U$ has an even order and thus a nontrivial Sylow 2-subgroup, and at the same time all its subgroups are cyclic. Hence, no complete mappings exist, providing an affirmative answer to the question.
A: OP asked me to fill in the details of my comment, and in attempting to do so I realised that I claimed too much. However, a very similar argument proves a weaker result which is strong enough to support Max Alekseyev's answer (or Fedor Petrov's comment to it).
Theorem: if there is a counterexample, the smallest $n$ which is a counterexample is prime.
Suppose to the contrary that there is a counterexample and that the smallest counterexample is $n = ab$ with $a, b > 1$. We note that the non-vanishing of the determinant encodes the property that $k \to k \pi_k \bmod n$ is a permutation. The multiples of $a$ map to multiples of $a$; the values of $k$ for which $a | \pi_k$ also map to multiples of $a$. Therefore to preserve the bijective nature of the mapping, we require $\pi$ to map multiples of $a$ to multiples of $a$.
Considering the restriction of $\pi$ to multiples of $a$ we can define $\sigma_k = \frac{\pi_{ak}}{a}$ which must be a permutation of $[b]$. But then $k \to k \sigma_k$ is also a permutation of $[b]$, so we have a smaller counterexample, contradicting our assumption. $\blacksquare$
