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][1] is more general; actually, I am mostly interested in $\mathbb{C}[x,y]$, so I asked the question above.
(2) See also [this question][2] (unfortunately, the fixed point theorems mentioned there are not relevant for $\mathbb{C}[x,y]$, but maybe there are generalizations of them that are relevant?).


Thank you very much!


  [1]: https://math.stackexchange.com/questions/2308771/is-it-possible-to-find-a-k-algebra-endomorphism-which-fixes-an-arbitrary-eleme/2308785#comment4750716_2308785
  [2]: https://mathoverflow.net/questions/105883/can-we-actually-find-any-fixed-points-with-brouwers-theorem?rq=1