Let $u,v \in \mathbb{C}[x,y]$. Let $\beta: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$ be the involution (= $\mathbb{C}$-algebra automorphism of order two) defined by $(x,y) \mapsto (x,-y)$.
Denote the symmetric elements with respect to $\beta$ by $S_{\beta}$ and the skew-symmetric elements w.r.t. $\beta$ by $K_{\beta}$.
Assume that $\mathbb{C}(u,v)$ is invariant under $\beta$, namely, $\beta(u) \in \mathbb{C}(u,v)$ and $\beta(v) \in \mathbb{C}(u,v)$.
Is it true that the following property P is satisfied: There exists an automorphism $g$ of $\mathbb{C}[x,y]$ such that $g(u),g(v) \in S_{\beta} \cup K_{\beta}$?
Example: $u=x+y^2$, $v=x+y+y^2$. It is easy to see that $\mathbb{C}(u,v)=\mathbb{C}(x,y)$ which is $\beta$-invariant. $u \in S_{\beta} \subset S_{\beta} \cup K_{\beta}$, $v \notin S_{\beta} \cup K_{\beta}$. However, there exists an automorphism $g$ of $\mathbb{C}[x,y]$ for which $g(u),g(v) \in S_{\beta} \cup K_{\beta}$: $g: (x,y) \mapsto (x-(y-x)^2,y-x)$. Indeed, $g(u)=g(x+y^2)=g(x)+g(y)^2=x-(y-x)^2+(y-x)^2=x \in S_{\beta}$ and $g(v)=g(x+y+y^2)=g(u+y)=g(u)+g(y)=x+(y-x)=y \in K_{\beta}$.
More generally: If $h: (x,y) \mapsto (u,v)$ is an automorphism of $\mathbb{C}[x,y]$, then $\mathbb{C}[x,y]=\mathbb{C}[u,v]$, so $\mathbb{C}(x,y)=\mathbb{C}(u,v)$ is $\beta$-invariant, and taking $g:=h^{-1}$ we obtain $g(u)=h^{-1}(u)=h^{-1}(h(x))=x \in S_{\beta}$ and $g(v)=h^{-1}(v)=h^{-1}(h(y))=y \in K_{\beta}$.
Therefore, a partial answer is: If $(x,y) \mapsto (u,v)$ is an automorphism of $\mathbb{C}[x,y]$, then property P is satisfied.
Remark: If I am not wrong, $\mathbb{C}(u,v)=\mathbb{C}(s_1,k_1,s_2,k_2)$, where $u=s_1+k_1, v=s_2+k_2, s_i \in S_{\beta}, k_i \in K_{\beta}$, $1 \leq i \leq 2$.
Any comments and hints are welcome!
Now asked in MO.