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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?

Added on 30 May 2020:

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

Rodrigo de Azevedo
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    This question has been already answered in MO. See https://mathoverflow.net/questions/361871/the-number-of-unitary-circulant-matrices-over-a-finite-field-mathbbf-q2 – Groups Jun 02 '20 at 06:39

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