Let $q=p^s$ be a prime power congruent to $1$ modulo $4$, let $\mathbb{F}_q$ be the finite field with $q$ elements, and let $\phi$ denote the Frobenius automorphism (that is $\phi(a)=a^p$ for every $a\in\mathbb{F}_q$).
Let $i$ be a primitive $4$-th root of unity in $\mathbb{F}_q$. Let $P=\mathbb{F}^{\times}_q/\langle i\rangle =\left\{\{a,ia,-a,-ia\}\mid a\in\mathbb{F}^{\times}_q\right\}$. Since $i^p=\pm i$ (depending on whether $p\equiv 1\pmod 4$ or $p\equiv 3\pmod 4$), the Frobenius automorphism acts on $P$.
I am interested in the number of orbits of $\phi$ on $P$.
There is a very nice, short unpublished note by Gareth Jones, Darryl Mccullough and Marcus Wanderley in which the number of orbits of $\phi$ acting on $\mathbb{F}_q$ is computed using very basic ideas (the Cauchy-Frobenius-Burnside lemma + basic facts from finite field theory), an ideal answer for me would be in that spirit.
It is easy to directly compute this number in the case $q=p^2$, it is:
$\frac{p^2+2p-3}{8}$ for $p\equiv 1\pmod 4$, and $\frac{p^2+4p-5}{8}$ for $p\equiv 3\pmod 4$.
Also, the case $q=p^k$, $k$ odd is easy (in this case we must have $p\equiv 1\pmod 4$) because we just divide by $4$ the number (minus $1$) computed in G. Jones et al.
I tried to figure out how to possibly combine the above results to obtain a general formula. No luck yet.