Describe the group of $\operatorname{Aut}_{K(x^p, y^q)}(K(x, y))$ Let $K$ be a field, $\operatorname{char}K = p > 0$, $q$ is prime such that $p-1 \equiv 0 \pmod{q}$.
Describe the group of $\operatorname{Aut}_{K(x^p, y^q)}(K(x, y))$.
It's pretty clear that we need to find a subgroup $G$ of $\operatorname{Aut}(K(x, y))$ such that $g(a) = a$,  $\forall a \in {K(x^p, y^q)}$, $\forall g \in G$, but how can I approach this task?

1) Since any automorphism fixing $x^p$ will fix $x$ (since  $p$-th roots are unique over fields of characteristic $p$) we are reduced to calculating $\operatorname{Aut}_{K(x, y^q)}(K(x, y))$ or, equivalently, $\operatorname{Aut}_{L(y^q)}(L(y))$ with $L=K(x)$.   
2) The required automorphisms are of the form $y\mapsto \zeta y$ where $\zeta\in L$ is a $q$-th root of unity.
So the problem boils down to calculating the group $\mu_q(L)$ of $q$-th roots of unity in $L$.
This group is the same as $\mu_q(K)$ and finally $$\operatorname{Aut}_{K(x^p, y^q)}(K(x, y))\simeq\mu_q(K)$$ 
3) If $K=\mathbb F_p$  the cardinality of $\mu_q(\mathbb F_p)$ is $gcd(p-1,q)$, which is $q$ by hypothesis.
Since $K$ contains $\mathbb F_p$ and since of course $K$ cannot contain more than $q$ elements we see that $\mu_q(L)=\mu_q(K)=\mu_q(\mathbb F_p)$ has $q$ elements.  
4) Conclusion
The group $\operatorname{Aut}_{K(x^p, y^q)}(K(x, y))$ consists of the $q$ automorphisms characterized by $$x\mapsto x, y\mapsto \zeta y \operatorname {with} \zeta \in \mu_q(K)$$
