Show $F(U) = K((x^q -x)^{q-1})$. Let $K$ be a finite field with $q$ elements. 
Show that if $U$ is the subgroup of $Aut(K(x)/K)$ which consists of all mappings $\sigma$ of the form $(\sigma \theta)(x) = \theta(ax+b)$ with $a \neq 0$ then $F(U) = K((x^q -x)^{q-1})$.
I am not getting any clue to solve the problem. Help Needed.
Here $F(U)$ is the fixed field of $U$. $K(X)$ is the field of fractions.
The elements of $K((x^q -x)^{q-1})$ are of the form $$\frac{f((x^q -x)^{q-1})}{g((x^q -x)^{q-1})}.$$
$F(U) = \{ \frac fg \in K(x) \mid \sigma(\frac fg) = \frac fg  \ \ \ \forall \sigma \in U\}$.
Hints are also welcome. Thank You.
 A: Lemma: Let $K$ be a field, and  $G\subseteq Aut (K(x)/K)$ a finite subgroup of order  $n$. Then the polynomial
$$p(T):=T^n+a_{n-1}T^{n-1}+\ldots+a_1T+a_0 =\prod_{\sigma \in G}(T-\sigma x)$$
is defined over the fixed field $K(x)^G$, and for any  $i\in \{0,1,\ldots,n-1\}$ such that $a_i\notin K$, we have $K(x)^G=K(a_i).$
Proof: Note that the  $\pm a_i$'s are given by the elementary symmetric polynomials  (in $n$ variables) evaluated at
$(\sigma_1x,\ldots,\sigma_n x)$. Thus $\sigma a_i=a_i$  for all  $\sigma \in G$, which gives $p(T)\in  K(x)^G[T ]$.  Also, since $x$ is transcendental over $K$ and 
$p(x)=0$, there must exist  $i\in \{0,1,\ldots,n-1\}$  such that $a_i\notin K$. For any such  $a_i$, we have  $a_i=g(x)/f(x)$, where $f, g\in K[x]$ are polynomials of degree $\leq n$.
Therefore, $g(T)-a_if(T) \in K(a_i)[T]$  is a polynomial of degree $\leq n$ vanishing at $x$, and then $[K(x):K(a_i)]\leq n$. Now, since   $K(a_i)\subseteq K(x)^G  \subseteq K(x)$, and $[K(x):K(x)^G]=|G|=n$  (Artin's thm), it follows that   $K(a_i)=K(x)^G$.         q.e.d.
Now, for  $K:=\mathbb{F}_q$  and $G:=\{x\mapsto ax+b|(a,b)\in K^*\times K\}\subseteq Aut(K(x)/K)$, we have 
$$p(T)=\prod\limits_{\sigma \in G}(T-\sigma(x))=\prod\limits_{(a,b)\in K^*\times K}(T-(ax+b)).$$  Since 
$$a_0:=p(0)=\prod\limits_{(a,b)\in K^*\times K}(-(ax+b))=-(x^q-x)^{q-1} \notin K,$$ 
the previous Lemma gives $K(x)^G=K((x^q-x)^{q-1})$.
