Let $\Gamma =\{z\in\mathbb{C}:|z|=1\}$, and $X=C(\Gamma)$. The rotation semigroup $\{T(t)\}_{t\geq 0}$, is defined as $$T(t)f(z)=f(\mathrm{e}^{it}z),\quad f\in X.$$ $Z\in X$, s.t. for all $z\in\Gamma$, $Z(z)=z$. Then $T(t)Z(z)=Z(\mathrm{e}^{it}z)=\mathrm{e}^{it}z$. Or we can say that any complex number $z=\mathrm{e}^{ix}$ is mapped to $\mathrm{e}^{i(x+t)}$. My question is, does $T(t)$ map $Z$ to itself? As I think, it is only true whenever $t$ is a multiple of $2\pi$.
P.S. I just want to confirm, either I am right or not!