If the eigenfunctions of $T_{\phi}f:=f\circ\phi$ span $C(X)$, then $\overline{\{\phi^{n}:n\in\mathbb{Z}\}}^{\text{unif}}$ is compact

Question

Let $X$ be a compact Hausdorff space. We endow $C(X,X)$ with the topology of uniform convergence. Let $\phi\colon X\to X$ be a homeomorphism and consider the linear isomorphism $T_{\phi}\colon C(X)\to C(X)$ defined by $T_{\phi}(f):=f\circ\phi$. Suppose that the eigenfunctions of $T_{\phi}$ span $C(X)$, that is,

$$C(X)=\overline{\text{span}\bigcup_{\lambda\in\mathbb{C}}\ker(T-\lambda I)}.$$

I want to show that $\overline{\{\phi^{n}:n\in\mathbb{Z}\}}^{\text{unif}}$ is a compact group.

I think there are versions of Arzela-Ascoli that may help, but I am not sure about this. Any suggestions would be greatly appreciated.

EDIT: Apparently, according to this article, there may be a proof in theorem 1 of the following article:

K. Sakai and S. Horinouchi: On compact transformation groups with discrete spectrum. Sci. Rep. Kagoshima Univ., 33, 1-5 (1984).

Unfortunately, I could not find this article online.