Let's assume we have a discrete dynamical system with $M \subset \mathbb{R}^n$ and
$$x_t=f(x_{t-1}), \text{ where } f: M \rightarrow M \text{ continuous} $$
Let's further assume we know all its cyclic points, i.e. we know all
$$x^*=f^k(x^*) \quad \forall k$$.
Is this information enough to know the topological conjugacy class of the dynamical system?
If this information is not enough, would it be enough if we further know that the map only has finitely many cyclic points?
Best Answer
We cannot infer the topological conjugacy class of $f$ even if there only many finitely many cyclic points.
A counterexample is given by circle rotations $S^1 \rightarrow S^1$. If the rotation angle is irrational, the dynamical system has no cyclic points. But any two circle rotations of different angles are not topologically conjugate.
You may consult Wikipedia [1] for starters or chapter 7.1 in the textbook "Introduction to Dynamical Systems" by Brin and Stuck.
[1] https://en.wikipedia.org/wiki/Rotation_number