Set-up
Say that we have a group generated by a single element '$g$' i.e. $G = \{n g\}_{n\in\mathbb{Z}} $ which is acting continuously on a compact metric space $(X, d)$.
So given any point $x \in X$, I can talk about the orbit-closure of $x$ i.e. the set $O(x)$ = $\overline{\{h.x\}_{h \in G}}$ = $\overline{\{(ng).x\}_{n\in\mathbb{Z}}}$.
Now this time rather than letting the whole group $\{n.g\}_{n\in\mathbb{Z}}$ act on $(X, d)$, I only consider the subset $\{f(n).g\}_{n \in\mathbb{Z}}$ acting on the metric space. Here $f$ is some polynomial that has integer roots.
We can again consider the orbit closure $O_{f}(x) = \overline{\{(f(n)g).x\}}$.
$O(x)$ being closed in compact metric space is again compact and now we can talk about $G = \{n.g\}_{n\in\mathbb{Z}}$ acting on just $O(x)$.
Small Definition
A point $y \in O(x)$ is recurrent if $\forall$ open neighborhood V of y, $\exists n \in\mathbb{Z}$ such that $[(ng).V] \cap V \neq\emptyset$.
A point $y \in O(x)$ is polynomially recurrent if $\forall$ open neighborhood V of y, $\exists n \in\mathbb{Z}$ such that $[(f(n)g).V] \cap V \neq\emptyset$.
Question
If we knew that the set of recurrent points in $X$ is dense, what could we say about set of polynomially recurrent points?
I tried taking $f(n) = n^{2}$ but then realized that integer squares are so negligible in integers that there might not be any recurrent points. And it is only going to get worse as the degree of the polynomial $f$ increases.
However, I do not seem to know how to answer my question one way or the another. I would really appreciate a help/hint/reference at this point.