$|z|=1$ is the unit circle in the complex plane.
Suppose $g$ is a self-homeomorphism of this circle of order $n$, $n \in \mathbb{N}$, and $g$ acts freely.
Is that true that $g$ must be defined by $z \mapsto e^{2\pi ik /n}z$ for some $k=1,...,n-1$.
Thank you in advance!
No. Let $f_k$ be your rotation, and let $g$ be literally any homeomorphism $S^1 \to S^1$. Then $gf_kg^{-1}$ is also an order $n$ homeomorphism; it's easy to see you can pick $g$ so that this new homeomorphism doesn't, e.g., send $1$ to $e^{2\pi ik/n}$.
On the other hand, every free homeomorphism of order $n$ is conjugate to one of yours. This follows from a classification of degree $n$ coverings of $S^1$.