Sketch the graph of $F_4^n(x)$ on the unit interval, where $F_4(x) = 4x(1-x)$. Conclude that $F_4$ has at least $2^n$ periodic points of period $n$. Sketching the graph I found that $F_4$ has exactly $2^{n-1}$ point wich prime period $n$. My $F_4^n$ graph look like a $|sin(x)|$ graph with $2^n$ intersections on $Id$, but I removed the $2^{n-1}$ to remove points that have period less then n. What is wrong?

Incomplete and with typos ... but contains a few hints and references.

Given the following functions:
$$\varphi(x)=-4x+2,\varphi^{-1}(x)=\frac{x-2}{-4}, g(x)=x^2-2$$
we have:
$$F_4(x)=(\varphi^{-1} \circ g\circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ g^{\circ n} \circ \varphi)(x)$$
Further to this, function $\gamma(x)=x+\frac{1}{x}$ has the property that
$$(g \circ \gamma)(x)=x^2+\frac{1}{x^2}=\gamma(x^2)\Rightarrow (\gamma^{-1} \circ g \circ \gamma)(x)=x^2\Rightarrow \\g(x)=(\gamma \circ t \circ \gamma^{-1})(x), t(x)=x^2$$
As a result
$$F_4(x)=(\varphi^{-1} \circ \gamma \circ t \circ \gamma^{-1} \circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ \gamma \circ t^{\circ n} \circ \gamma^{-1} \circ \varphi)(x) \tag{1}$$
Effectively, by introducing $\gamma$ we have to exclude $x=0$ perioding point, which is not of period $n$ though.
Periodic points of $t(x)=x^{2}$ are periodic for $F_4(x)$ "through" $\gamma(x)$ and $\varphi(x)$. Complex roots of $x^{2^n-1}=1$ are periodic points for $t(x)$ (and  $F_4(x)$ "through" $\gamma(x)$ and $\varphi(x)$) and, except $x=1$, all the other roots have period $n$ (there is an entire chapter in  this book, starting with page 6, dedicated to this function). To see all these working, let's take $x=\sin^2{\frac{\alpha}{2}} \in (0,1)$ then (I will omit the details as those are easy to check)
$$\varphi(x)=2\cos{\alpha}$$
$$\gamma^{-1}\left(2\cos{\alpha}\right)=e^{i\alpha}$$
$$t^{\circ n}\left(e^{i\alpha}\right)=e^{i2^{n}\alpha}$$
$$\gamma\left(e^{i2^{n}\alpha}\right)=2\cos{\left(2^{n}\alpha\right)}$$
$$\varphi^{-1}\left(2\cos{\left(2^{n}\alpha\right)}\right)=\sin^2{\left(2^{n-1}\alpha\right)}$$
or
$$F_4^{\circ n}\left(\sin^2{\frac{\alpha}{2}}\right)=\sin^2{\left(2^{n-1}\alpha\right)} \tag{2}$$
Complex roots of $z^{2^n-1}=1$ are of the form
$$z_k=\exp{\left(\frac{2i\pi k}{2^n-1}\right)},k=\overline{0..2^n-1}$$
thus taking $$\alpha=\frac{2\pi k}{2^n-1}$$
we have
$$F_4^{\circ n}\left(\sin^2{\frac{\pi k}{2^n-1}}\right)=\sin^2{\left(2^{n-1}\frac{2\pi k}{2^n-1}\right)}=\sin^2{\left(\frac{(2^{n}-1+1)\pi k}{2^n-1}\right)}=\\
\sin^2{\left(k\pi + \frac{\pi k}{2^n-1}\right)}=\sin^2{\frac{\pi k}{2^n-1}},k=\overline{0..2^n-1} \tag{3}$$
Note: we could simply apply (from $(2)$) $\sin^2{\frac{\alpha}{2}}=\sin^2{\left(2^{n-1}\alpha\right)} \Rightarrow \frac{\alpha}{2}+k\pi=2^{n-1}\alpha$ and arrive to the same conclusion, but without conjugacy functions reaching this point would require having a good intuition.
However
$$F_4^{\circ n}\left(\sin^2{\frac{\pi (2^n-k)}{2^n-1}}\right)=\sin^2{\left(\frac{\pi (2^n-k)}{2^n-1}\right)}=\sin^2{\left(\frac{\pi (2^n-1-k+1)}{2^n-1}\right)}=\\
\sin^2{\left(\pi - \frac{\pi (k-1)}{2^n-1}\right)}=\sin^2{\left(\frac{\pi (k-1)}{2^n-1}\right)}=F_4^{\circ n}\left(\sin^2{\left(\frac{\pi (k-1)}{2^n-1}\right)}\right) \tag{4}$$
this last statement reduces the number of periodic points with period $n$ to $2^{n-1}$.
Note: $(4)$ can also simply be obtained from $\frac{2\pi k}{2^n-1}=\frac{2\pi m}{2^n-1} + p\pi$, where $0\leq k,m \leq 2^n-1$ and concluding that $p=1$ and $k-m=2^n-1$.
