A $\mathbb{C}$-linear automorphism of order $2$ on $\mathbb{C}[x,y]$, which does not preserve multiplication

Let $f: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$ be a $\mathbb{C}$-linear automorphism of order $2$, thinking of $\mathbb{C}[x,y]$ as a $\mathbb{C}$-vector space (forgetting about the multiplication). Denote the natural basis over $\mathbb{C}$ by $b_{ij}:= \{x^iy^j \}$.

If I am not wrong, $\{f(b_{ij})\}$ is also a $\mathbb{C}$-basis (for a finite dimensional vector space, the images of a basis elements under a linear isomorphism form a basis; I guess this is also true in the infinite dimensional case).

Is it possible to find such $f$ which does not preserve the multiplication in $\mathbb{C}[x,y]$? In other words, is it possible to find such $f$ which is not a ring homomorphism of $\mathbb{C}[x,y]$?

Of course, if $f$ preserves the ring multiplication, then it is a ring (algebra) automorphism.

Any hints are welcome!

Sure there is such an $f$. For instance, there is the map defined by adding $1$ to all even exponents and subtracting $1$ from all odd exponents. So $1=x^0y^0\mapsto xy$ and $x^3y^6\mapsto x^2y^7$, and so on.
• Thanks, but in your example $1 \mapsto xy$, and I wish that $f$ will be the identity on $\mathbb{C}$. – user237522 Apr 30 '17 at 0:43
• Well, ok. Let all $0$ exponents be untouched, but otherwise add $1$ to all odd exponents and subtract $1$ from all even exponents (the other way around from what is in the answer). So $x^5=x^5y^0\mapsto x^6y^0$, and $xy\mapsto x^2y^2$. – Arthur Apr 30 '17 at 0:47
• Nice example, thanks! Do you think that there is any 'weak' additional condition which guarantees that $f$ preserves the multiplication? – user237522 Apr 30 '17 at 1:04