Let $f,g \in F [t]$ be relatively prime and suppose that $u = f/g$ lies in $F (t) - F $. Prove that $G (F (t)/F) $ consists of all $F $-automorphisms of $F (t)$ mapping $t $ to $(at+b)/(ct+d) $ where $a,b,c,d \in F $ satisfies $ad-bc \neq 0$.
The above should use the following result (proof of which is skipped):
$F (t) / F(u) $ is finite of degree $d=max \{ deg (f), deg (g) \} $ with the minimal polynomial $f-ug $.
The image of $t $ completely determines all elements of $G $. If $t $ gets mapped to some $u=f/g $ then by the above it should be that (after degree consideration) $deg f = deg g = 1$.
The only question I wish to ask is: why can $u $ be not a constant in $F $, which is equivalent to the last condition that $ad-bc=0$?
If $t$ were mapped to some constant $u \in F$, note that $u$ is also mapped to $u$ (since the automorphism is supposed to fix $F$), thus the mapping is not a bijection and thus cannot be an automorphism.