Let $u,v,w \in \mathbb{C}(x,y)-\mathbb{C}$.
Is it possible to find an automorphism $f$ of $\mathbb{C}(x,y)$ that fixes $u$ and $v$, and not fixes $w$? namely, $f(u)=u, f(v)=v, f(w)\neq w$.
Perhaps this question is relevant.
Remark: If $\mathbb{C}(u,v) \subseteq \mathbb{C}(u,v)(w)=\mathbb{C}(x,y)$ is Galois of order $>1$, then my question has a positive answer: One can take any $f \neq 1$ from the Galois group.
Any comments are welcome!
The answer is no in general.
For example, if $w=u+v$ then $f(w)=f(u)+f(v)=u+v=w$.
Same way, since $f$ fixes $\mathbb Q$, any $w \in \mbox{Span}_{\mathbb Q} \{ 1, u, v \}$ is a counter example.
You might need to throw in the condition that $1,u, v, w$ are linearly independent over $\mathbb Q$ or some algebraic independence condition, if you want to get a positive result.
Edit: As Alex pointed, linear independence is not enough. For example if $w=P(u,v)$ for some polynomial $P$, then $f(u)=u, f(v)=v$ implies $f(w)=w$.