$\def\CC{\mathbb{C}}$Let $K = \CC(x\_1, \ldots, x\_n)$ and let $G$ be a countable group of automorphisms of $K$; in the cases I care about, $G \cong \mathbb{Z}$. Then the field of $G$-invariants, $K^G$, is an extension of $\CC$ of some transendence degree $d$. I would like to know under what circumstances I can say that the Zariski closure of the $G$-orbit of a generic $n$-tuple has dimension $n-d$.


I want to be a little sloppy about what I mean by generic, but I don't think there is anything deep in that issue. For "generic" $x \in \CC^n$, all the rational maps $g$ are defined at $x$, so we can consider $\{ g(x) \}\_{g \in G} \subset \CC^n$ and take its Zariski closure. Every function in $K^G$ is constant on this Zariski closure, so the Zariski closure has dimension $\leq n-d$.


**Question** Can I conclude that the dimension is "generically" $n-d$?



Dimension of orbit versus invariant functions
Yes for $G=\mathbb Z$, see Theorem 4.1 of this paper by [Amerik-Campana](https://arxiv.org/pdf/math/0510299.pdf).


