Group of automorphisms of field of rational functions over finite field

Question

I'm trying to solve the following problem but am not sure of the best way to start or the best strategy to use. Let $F$ be a finite field with $q$ elements, and let $K = F(x)$ be the field of rational functions over $F$. Let $G$ be the group of all automorphisms $\sigma$ of $K$ such that $$\sigma(x) = \frac{ax + b}{cx + d}$$ where $a, b, c, d \in F$ and $ad - bc\neq 0$.

(1) Show that the order of $G$ is $q^3 - q$.

(2) Show that the fixed field $K^G$ of $G$ is $F(y)$, where $$y = \frac{(x^{q^2} - x)^{q+1}}{(x^q - x)^{q^2 + 1}}.$$

Any hints or strategies would be appreciated!

Answer 1

Since $F$ has $q$ elements there's $q^4$ choices for values of $a$, $b$, $c$, and $d$. Therefore $|G| \leq q^4$.

The requirement that $ad-bc \neq 0$ along with counting the number of values of $a$, $b$, $c$, and $d$ that give equivalent $\sigma$'s reduces $|G|$ below $q^4$. Try to start counting those possibilities.