Let $G$ be a finite group. Given complex numbers $x=\{x_g: g\in G\}$, one can define a $|G|\times |G|$ matrix $X$, with entries $X_{g,h} = x_{gh^{-1}}$.
Let's consider $G$ being the symmetric group $S_n$. My question is whether there is some simple way to show the matrix $X$ is invertible (i.e. its determinant is non-zero) for a given $x$, without checking all irreducible representations.
To be specific, consider the following example. The determinant of such a matrix $X$ was calculated in a paper by Zagier, with $$ x_g = q^{n-k}, \quad \text{if } g=T_{k,n} $$ for $k=1,\ldots,n$, and $x_g=0$ otherwise. Here $q\in \mathbb{C}$ is a fixed number, $T_{k,n} \in S_n$ is the cyclic shift on the last $n-k+1$ elements that moves the $k$th element to the last position. Using the language of group algebra and by factorizing the element $\sum_{k=1}^n q^{n-k} T_{k,n}$, Zagier showed (Thm 2' therein) that $\det(X)=\prod_{k=1}^{n-1} (1-q^{k(k+1)})^{\frac{n!}{k(k+1)}}$. In particular, $\det X$ is non-zero if $q$ is not a root of unity.
My question is whether there is a cheaper way to prove $\det X$ is non-zero. Specifically, I am considering the following situation: $$ x_g = n-k, \quad \text{if } g=T_{k,n}. $$ In this case it seems Zagier's clever factorization technique no longer works.
