Let $w \in \mathbb{C}[x,y]-\mathbb{C}$ and let $u \in \mathbb{C}[x,y]-\mathbb{C}[w]$.

Is it possible to find a $\mathbb{C}$-algebra endomorphism $f$ of $\mathbb{C}[x,y]$ such that $f(w)=w$ and $f(u) \neq u$?

There are special cases having a positive answer, for example: $w=x^2+y^2$, $u=x$; in this case, one can take $f=\alpha: (x,y) \mapsto (y,x)$ the exchange involution (more generally, if $w$ is symmetric with respect to some involution $\iota$ on $\mathbb{C}[x,y]$, and $u$ is non-symmetric with respect to that involution $\iota$, then $\iota(w)=w$ and $\iota(u) \neq u$). However, in my question there is no such information about $w$ and $u$.

Is it hopeless to try to find such $f$ or perhaps it is possible to apply one of the many fixed point theorems to solve my question in the affirmative?

Remarks: (1) This quesiton is more general; actually, I am mostly interested in $\mathbb{C}[x,y]$, so I asked the question above. (2) See also this question.

Thank you very much!