I asked this question in [MSE](https://math.stackexchange.com/questions/3694921/the-number-of-unitary-circulant-matrices-over-a-finite-field-mathbbf-q2) few days ago but there was no response.


Suppose $\mathbb{F}=\mathbb{F}\_{q^2}$, where $q$ is a prime power. The conjugate of elements in $\mathbb{F}$ is defined by $\overline{x}=x^q$. I need to find the number of $n\times n$ unitary circulant matrices over $\mathbb{F}$.


The number of invertible circulant matrices over a finite field can be seen elsewhere, such as [when $n,q$ coprime](https://math.stackexchange.com/questions/2363372/number-of-invertible-circulant-matrices-over-a-finite-field) and my question [when $n=\operatorname{char} q$](https://math.stackexchange.com/questions/3646138/number-of-invertible-elements-in-mathbbf-qx-langle-xp-1-rangle-with-p).


Is there any better method to calculate this number other than considering each entry?


This is equivalent to the order of the centraliser of the permutation matrix of $(1,2,\dots,n)$ in $\operatorname{GU}\_n(q)$.




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**Added on 30 May 2020 MSE:**


Let $C$ be the subgroup of $\operatorname{GL}\_n(q^2)$ of all circulant matrices. Is $C\operatorname{GU}\_n(q)$ a subgroup of $\operatorname{GL}\_n(q^2)$? That is, is $C\operatorname{GU}\_n(q)=\operatorname{GU}\_n(q)C$? If that is correct then $C\operatorname{GU}\_n(q)=\operatorname{GL}\_n(q^2)$ and so $|C\cap\operatorname{GU}\_n(q)|$ follows. Here we denote by $\operatorname{GU}\_n(q)$ the general unitary group over $\mathbb{F}\_{q^2}$.



The number of unitary circulant matrices over a finite field $\mathbb{F}_{q^2}$
Let $\tau$ denote the permutation matrix corresponding to $(1,2,\ldots,n)$. Consider it first as an element of $\mathrm{M}\_n(q^2)$. 


This matrix has minimal polynomial equal to $X^n-1$, which is equal to its characteristic polynomial. It is therefore cyclic, and its centralizer is isomorphic to $\mathbb{F}\_{q^2}$-algebra $\mathbb{F}\_{q^2}[X]/(X^n-1)$. *For simplicity, I'll assume $\mathbb{F}\_{q^2}$ has no $N$-th root of unity, so that this algebra is isomorphic to $\mathbb{F}\_{q^{2n}}$.* Mapping $X\to \bar{X}^t$ defines a field automorphism of order $2$ of $\mathbb{F}\_{q^{2n}}$ which, by Hilbert 90 (which is an overkill, but does the job), restricts to a surjective map $\mathbb{F}\_{q^{2n}}^\times\to \mathbb{F}\_{q^n}^\times$. The centralizer you're seeking is precisely the kernel of this map, and is of carinality
$$ \frac{q^{2n}-1}{q^n-1}=q^n+1.$$


Does this make sense?


Now, if $X^N-1$ splits in $\mathbb{F}\_{q^2}$, then $\mathbb{F}\_{q^2}[X]/(X^N-1)$ is a product of finite fields, and the same type of argument works per coordinate, which should result in a formula of the form $\prod\_{i=1}^r q^{d\_i}+1$ for suitable $d\_i$'s such that $\sum d\_i=n$.


