I came across this while trying to do a problem frm ergodic theory. Let $X$ be a compact metric space, and let $f:X\rightarrow X$ be a continuous function. Can we say anything about the equicontinuity of $\{f^n\}_{n=1}^\infty$? ($f^n$ is the composition of $f$ with itself $n$ times).
I thought about showing that $\{f^n\}_{n=1}^\infty$ has compact closure, using a diagonal argument to extract (from a subsequence) another subsequence which converges pointwise on a countable dense set. However, I then got stuck. I also tried looking for a counterexample (I have no idea if the proposed result actually holds), but came up empty-handed.
The following is true:
If $(X,T)$ is distal, then $\{T^n\}$ is equicontinuous it'd there exists an equivalent metric on $X$ making it an isometry. The skew product on $\mathbb{T}^2$ given by
$(x,y) \mapsto (x+\alpha,y+x)$
Is distal but cannot be made into an isometry