Let $n\geq2$ be a natural number and let $x_1,\ldots,x_n$ be $n$ real numbers. Is there a general sufficient condition to guarantee that the set of $n$ "cyclicly permuted" vectors $\left\{(x_1,x_2,\ldots,x_n), (x_n, x_1,\ldots,x_{n-1}), \ldots, (x_2,x_3,\ldots,x_1)\right\}$ is linearly independent?
When $n=2$ it is sufficient and necessary that $(x_1+x_2)(x_1-x_2)\ne0$. When $n=3$, the determinant of the matrix whose rows are those vectors is $(x_1+x_2+x_3)\left(\frac{{(x_1-x_2)}^2+{(x_1-x_3)}^2+{(x_2-x_3)}^2}{2}\right)$, so the necessary and sufficient condition is that the sum of those $3$ numbers is not zero, and they are not all the same number.
In general, the determinant of the matrix is $$ \prod_{k=0}^{n-1}\left(\sum_{\ell=0}^{n-1}e^{\frac{2\pi ik\ell}{n}}\cdot x_{\ell+1}\right). $$ But I am unable to deduce some intuitive condition for that to be non-zero.
Since I think that determinant is a product of discrete Fourier transforms, I also tag this as related to Fourier transform.
If this is inappropriate, I will remove that tag.
P.S.
The determinant can be found as Lemma 5.26 in Washington's Introduction to cyclotomic fields.
Any help is greatly appreciated.
A sufficient condition is that one of the elements dominates the others. Namely, for some $i\in[n]$, $|x_i|> \sum_{j\neq i}|x_{j}|$. Note that we can choose the matrix to have only $x_i$ on its diagonal entries. Then, our matrix is strictly diagonally dominant, making it non-singular.