Suppose we have a bijective function $f: \mathbb{Z}[i] \rightarrow \mathbb{Z}[i]$ over the Gaussian integers. I want to know if there exists a polynomial $g(x)$ with coefficients also in $\mathbb{Z}[i]$ so that $\forall \beta \in \mathbb{Z}[i], \; f(\beta)=g(\beta)$.
I'm in the middle of my undergraduate Galois theory course, but this question is for a software project outside of that class. The idea is to represent transformations on a grid of characters (in an ASCII game similar to Rouge or Nethack) via compositions of functions that are quick to evaluate, and require very little memory to map a potentially very large number (maybe even arbitrary number) of characters to specific locations. The idea is inspired by how projection and position matrices are composed in 3D graphics pipelines for calculating the positions of vertices on a screen.
What I've tried:
Any polynomial $g(x)$ that is a bijective function under evaluation at arbitrary $\beta$, must have the property of having only one root $\alpha \in\mathbb{Z}[i]$, which means $g(x)=0 \implies x=\alpha$. We then know $(x-\alpha)\; | \;g(x)$ but $g(x)=(x-a)h(x)$ should mean $h(x) $ is irreducible in $\mathbb{Z}[i]$ and has degree $>1$ since there are no other roots.
I also know that $$f(\beta)=\sum_{i=0}^n g_i \: \beta^i=(\beta-\alpha)h(\beta)$$
But that's where I get stuck. I have a feeling that such a polynomial doesn't exist, but I'm not sure how to proceed with finding a counterexample $f$ because I'm not sure how I could construct such a polynomial to match $f$, or prove that one does not exist.
What about the function that swaps $0$ and $1$ and leaves everything else fixed? That can't be a polynomial since it has infinitely many fixed points.
This argument works over any infinite field.