First appearance degree conjecture for standard circulant matrices

Write $T_{n-1}=n(n-1)/2$, and define

$$\operatorname{APD}_m(A) =\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \left(\sum_{j=1}^nA_{j,\sigma(j)}\right)^m.$$

The determinant expansion gives the exponential generating function

$$\sum_{m\ge0}\operatorname{APD}_m(A)\frac{z^m}{m!} =\det\big(e^{zA_{ij}}\big)_{1\le i,j\le n}.$$

For the standard circulant

$$(C_n)_{ij}=((i+j-2)\bmod n)+1,$$

put $y=e^z$. Reversing rows $2,\ldots,n$ converts $(y^{(C_n)_{ij}})$ into the usual right-circulant with first row $(y,y^2,\ldots,y^n)$, contributing the sign $(-1)^{(n-1)(n-2)/2}$. If $\omega$ is a primitive $n$-th root of unity, its circulant eigenvalues are

$$\sum_{j=0}^{n-1}y^{j+1}\omega^{kj} =\frac{y(1-y^n)}{1-y\omega^k} \qquad(0\le k<n).$$

Since

$$\prod_{k=0}^{n-1}(1-y\omega^k)=1-y^n,$$

we obtain

$$\sum_{m\ge0}\operatorname{APD}_m(C_n)\frac{z^m}{m!} =(-1)^{(n-1)(n-2)/2}e^{nz}(1-e^{nz})^{n-1}.$$

This has a zero of exact order $n-1$ at $z=0$. Therefore

$$m_1(C_n)=n-1,$$

and its first nonzero coefficient is

$$\operatorname{APD}_{n-1}(C_n) =(-1)^{n(n-1)/2}n^{n-1}(n-1)! =(-1)^{T_{n-1}}n^{n-2}n!.$$

For the fixed-point statistic, the same determinant identity gives

$$\sum_{m\ge0}\operatorname{APD}_m(\operatorname{fix})\frac{z^m}{m!} =\det\big(J_n+(e^z-1)I_n\big) =(e^z-1)^{n-1}(e^z+n-1).$$

Consequently

$$m_1(\operatorname{fix})=n-1, \qquad \operatorname{APD}_{n-1}(\operatorname{fix})=n!.$$

This proves both asserted comparisons. More generally, the complete closed formula is

$$\operatorname{APD}_m(C_n) =(-1)^{T_{n-1}}n^m(n-1)!\,{m+1\brace n},$$

where ${m+1\brace n}$ denotes a Stirling number of the second kind.