Let $S = \{z \in \mathbb{C} : |z| = 1\}$ and define the doubling map $f : S \to S$, $f(z) = z^2$. Its iterates are $f^n(z) = z^{2^n}$. Let $X$ be the set of all points on the unit circle that are $2^n$th roots of unity, for some $n \ge 1$.
I don't know how to show if the map $f$ restricted to $X$ has dense positive semiorbits or not, i.e if there is $z \in X$ such that $\mathcal{O^+}(z) = \{f^k(z) : k =0, 1, 2, ...\}$ is dense. Can someone help me?
P.S. Has $f$ restricted to $X$ a dense orbit? (i.e. is there $z \in X$ such that $\mathcal{O}(z) = \{f^k(z) : k \in \mathbb{Z}\}$ is dense?)
The answer to your question is no, because the semiorbit of any point of $X$ under $f$ is finite.
To see this, note that by definition any $2^n$-th root of unity $\zeta$ satisfies $\zeta^{2^n} = 1$. Thus the semiorbit of $\zeta$ contains at most $n$ points: $$ \zeta, \zeta^2,\dotsc, \zeta^{2^{n-1}}, \zeta^{2^n} = 1, 1, \dotsc $$
Edit: As for the postscript, I claim that $\mathcal{O}^-(1) = X$. To prove this it's enough to show that $$ f^{-n}(1) = \left\{ z \in \Bbb{C}: z^{2^n} = 1\right\} =: \mu_n $$ since $X = \bigcup_{n \geq 1} \mu_n$. But it is clear from the definition of $\mu_n$ that $f^n(\mu_n) = \{1\}$, while if $z \in f^{-n}(1)$, then $f^n(z) = z^{2^n} = 1$, so $z \in \mu_n$.
Update: In the comments, the OP asked whether $X$ has any isolated points. The answer is no. To see why, note that the $2^n$ points of $\mu_n$ are equally spaced on the unit circle. Furthermore, to obtain the points of $\mu_{n+1}$ all you need to do is add the middle point of any arc between consecutive points of $\mu_n$.
For a more formal proof, observe that any element of $\mu_n$ can be written as $$ e^{2\pi i k/2^n} $$ for some $0 \leq k < 2^n$. Thus the length of the arc of unit circle between any two consecutive elements is $\pi/2^{n-1}$. In particular, fix a point of $x \in X$ and some $\varepsilon > 0$. If $m > \log 2 \cdot \log \frac{\pi}{\varepsilon} + 1$, then you can find a point $y \in \mu_m \subset X$ such that $\lvert x - y \rvert < \varepsilon$.