let G be the subset of $\operatorname{aut}_{K} K(x)$ consisting of the three automorphisms
$ x \mapsto x $ $ x \mapsto 1/(1-x)$ $ x \mapsto (x-1)/x$
then G is a subgroup of $\operatorname{aut}_K K(x)$. determine the fixed field of $G$
solution:(wrong) let $ f/g \in K(x)$ with $f$ and $g$ relatively prime in K[x], and suppose that $f/g$ is in the fixed field then $ f/g = 1/(1-(f/g))$ which gives $ f^2 -fg=g^2$
$g^2 = f(g-f)$ so we have that $ f \mid g^2$ so we must have that $f$ is a constant since $f$ and $g$ are relatively prime. and by symmetry we have that $g$ must be a constant which is a contradiction so we must have that $f/g$ is not in the fixed field of G, but this is true for every $f/g \in $ K(x) so the fixed field of G must be empty.
is that right im not sure if that is a contradiction or not?
new solution:
let $\dfrac{ax+b}{cx+d} \in Aut_{K}K(x)$ $ \sigma_{1} :x \mapsto x $ $ \sigma_{2} :x \mapsto 1/(1-x)$ $ \sigma_{3} :x \mapsto (x-1)/x$
$\dfrac{ax+b}{cx+d} = x$ gives $ax+b=cx^2+dx$, which gives $c=0$, $b=0$, and $a=d$ or $(a,b,c,d)= (a,0,0,a)$
$\dfrac{ax+b}{cx+d} =1/(1-x)$ gives $ax+b-ax^2-bx =cx+d$ , gives $a=0$, $b=d$, and $c=-b$ or $(a,b,c,d)=(0,b,-b,b)$
$\dfrac{ax+b}{cx+d} = (x-1)/x$ gives $ax^2+bx=cx^2+dx-cx-d$ gives $a=c$, $d=0$, $b=-c$ or $(a,b,c,d)= (a,-a,a,0)$