I asked this question in MSE 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 and my question when $n=\operatorname{char} q$.

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 $\operatorname{GU}_n(q)$.

**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 $\mathbb{F}_{q^2}$.