Element structure of general semilinear group of degree one over a finite field

This article describes the element structure of the general semilinear group of degree one. Recall that, for a field $$K$$, the group is defined as:

$$\Gamma L (1,K) = GL(1,K) \rtimes \operatorname{Gal}(K/k) = K^\ast \rtimes \operatorname{Gal}(K/k) $$

where $$GL(1,K) = K^\ast$$ is the multiplicative group of $$K$$, $$k$$ is the prime subfield of $$K$$, and $$\operatorname{Gal}(K/k)$$ denotes the Galois group of $$K$$ over $$k$$.

We are interested specifically in the case where $$K$$ is a finite field of size $$q$$, in which case the group is written as $$\Gamma L(1,q)$$. Suppose $$q$$ is a prime power with underlying prime $$p$$, so that $$q = p^r$$ for a positive integer $$r$$. $$p$$ is the characteristic of $$K$$. In this case, $$K^\ast$$ is cyclic of order $$q - 1$$ (see multiplicative group of a finite field is cyclic) and $$\operatorname{Gal}(K/k)$$ is cyclic of order $$r$$ (generated by the Frobenius map $$a \mapsto a^p$$).

Thus, $$\Gamma L(1,K)$$ is a metacyclic group of order $$r(q - 1)$$ with presentation:

$$\langle a,x \mid a^q = a, x^r = e, xax^{-1} = a^p \rangle$$

(here $$e$$ denotes the identity element).

Note that if $$r = 1$$, the group is the same as $$GL(1,q) = \mathbb{F}_q^\ast$$ and is cyclic of order $$q - 1$$.

Number of conjugacy classes formulas
For every fixed value of $$r$$, the number of conjugacy classes simply becomes a polynomial in $$p$$. The values of these polynomials for small $$r$$ are listed below:

For generic r
This table is complicated! We consider conjugacy classes in the multiplicative group and conjugacy classes outside the multiplicative group. Note that $$q = p^r$$.

For r = 2
We take $$q = p^2$$:

For r = 3
We take $$q = p^3$$: