Let $K$ be an algebraically closed field of characteristic zero, and $X$ be an affine $K$-variety (identify $X$ with its set of $K$-points). Let $G$ be group acting "abstractly" on $X$, by which I mean there is simply a group homomorphism $\rho:G \to \operatorname{Aut}(X)$. Then $G$ also acts on the coordinate ring $K[X]$ by $K$-algebra automorphisms (there is an associated group homomorphism $\rho^*:G \to \operatorname{Aut}(K[X])$.
It is a classical fact that if $G$ is an algebraic $K$-group (identify $G$ with the group of points $G(K$)) acting algebraically on $X$ (i.e. the map $G \times X \to X$ is a morphism of varieties), then the orbits in $K[X]$ span finite-dimensional $K$-subspaces of $K[X]$.
Are there examples in which a group $G$ acts on an affine variety $X$ where this finite-dimensionality property for orbits in $K[X]$ holds for some other reason than the action being algebraic? In particular, I am interested in the case when $G$ is the elementary subgroup of a Chevalley group with coefficients in a ring of $S$-integers, something like $G = \operatorname{SL}_3(\mathbb{Z})$ or $G = \operatorname{SL}_3(\mathbb{Z}[\sqrt{d}])$.
So far, the only example I have is incredibly trivial, namely $X = \mathbb{A}^1_K$ is the affine line with coordinate ring $K[x]$. Since automorphisms of $K(x)$ are möbius transformations, $K$-algebra automorphisms of $K[x]$ are affine transformations. In particular, I believe this forces orbits to be finite-dimensional for any group acting on $K[x]$.
I tried extending this example to $X = \mathbb{A}^2_K$, using the result that automorphisms of $K[x,y]$ are tame. However, tameness is still not particularly close to being affine, even for automorphisms of finite order. I think this question is related.