I came across this question today and am looking for a flaw in this very simple argument of mine.
I'm interested in whether a function $f : \mathcal{X}^n \to \mathbb{R}_{\geq0}$ is radially symmetric. I have not done much previous work with radial symmetry but How do you prove that multiplying a symmetric function by a constant preserves symmetry? claims that a function $f$ is radially symmetric if there exists a function $g$ such that $f(x) = g(\lVert x \rVert)$.
Now, if the domain of $f$ if discrete, i.e. $\mathcal{X}$ is discrete, there exists an injective Gödel encoding $h : \mathcal{X}^n \to \mathbb{N}$ with inverse $h^{-1}$ such that $f = f \circ h^{-1} \circ h$. Wouldn't that yield that if $h$ is a norm which I suppose it is that $f$ is radially symmetric purely because it is defined on a discrete domain?
The statement seems wrong to me, do you see the error in the argument?