Let $n>1$ be odd. In my 2019 preprint On some determinants involving the tangent function, I proved that $$\det\left[\tan\pi\frac{aj+bk}n\right]_{1\le j,k\le n-1}=\left(\frac{-ab}n\right)n^{n-2}$$ for any integers $a$ and $b$ with $\gcd(ab,n)=1$, where $(\frac{\cdot}n)$ denotes the Jacobi symbol; in particular, $$\det\left[\tan\pi\frac{j-k}n\right]_{1\le j,k\le n-1}=n^{n-2}.$$
Note that $$\tan\pi x=\frac{2\sin\pi x}{2\cos\pi x}=i\frac{1-e^{2\pi ix}}{1+e^{2\pi i x}}.$$ So I actually evaluated the determinant $$\det\left[\frac{1-\zeta^{j-k}}{1+\zeta^{j-k}}\right]_{1\le j,k\le n-1}$$ with $\zeta=e^{2\pi i/n}$. For $j,k=1,\ldots,n-1$ let's define $$f(j,k)=\begin{cases}\cot\pi\frac{j-k}n&\text{if}\ j\not=k,\\0&\text{if}\ j=k.\end{cases}\label{1}\tag{$*$}$$ When $j\not=k$, we have $$f(j,k)=-i\frac{1+\zeta^{j-k}}{1-\zeta^{j-k}}$$ Thus \begin{align*}\det[f(j,k)]_{1\le j,k\le n-1}=&\sum_{\tau\in D(n-1)}\mathrm{sign}(\tau)(-i)^{n-1}\prod_{j=1}^{n-1}\frac{1+\zeta^{j-\tau(j)}}{1-\zeta^{j-\tau(j)}} \\=&(-1)^{(n-1)/2}\sum_{\tau\in D(n-1)}\mathrm{sign}(\tau)\prod_{j=1}^{n-1}\frac{1+\zeta^{j+\tau(j)}}{1-\zeta^{j-\tau(j)}}, \end{align*} where $$D(n-1)=\{\tau\in S_{n-1}:\ \tau(j)\not=j\ \ \text{for all}\ j=1,\ldots,n-1\}.$$
Motivated by the above and Question 403025 of mine, I have formulated the following new conjecture.
Conjecture. Let $n>1$ be an odd number. Then $$\sum_{\tau\in D(n-1)}\mathrm{sign}(\tau)\prod_{j=1}^{n-1}\frac{1+\zeta^{j-\tau(j)}}{1-\zeta^{j-\tau(j)}}=\frac{(-1)^{(n-1)/2}}n((n-2)!!)^2\tag{1}$$ for any primitive $n$-th root $\zeta$ of unity; in particular, $$\det[f(j,k)]_{1\le j,k\le n-1}=\frac{((n-2)!!)^2}n\tag{2}$$ with $f(j,k)$ given by \eqref{1}.
My numerical computaton supports this conjecture.
QUESTION. Is the conjecture true? Can one supply a proof?
