We have a compact metric space $(X,d)$ and a homeomorphism $T:X\to X$.
For any ultrafilter $p\in\beta\mathbb{Z}$ we can define the map $T^p:X\to X$
given by $T^p(x):=\lim_{n\to p}T^n x$ (which can very well be discontinuous in general).
We want to prove that $\{T^n\}$ is an equicontinuous family iff $T^p$ is a homeomorphism for every $p\in\beta\mathbb{Z}$.
The implication $(\Rightarrow)$ is easy, while I can't prove $(\Leftarrow)$.
So far, assuming that $T^p$ is a homeomorphism, I proved that $T^pT^q=T^qT^p$ for every $p,q\in\beta\mathbb{Z}$. Moreover, for any distinct $x\neq y$ we have $\inf_n d(T^n(x),T^n(y))>0$, but this is not enough to have equicontinuity.
Clearly it would suffice to know that the map $$ \beta\mathbb{Z}\times X\to X,\ \ \ (p,x)\mapsto T^p(x) $$ is jointly continuous, while we only know that it is separately continuous in $p$ and $x$.
Any ideas?