Question 21 Let $f(x)$ be a function on $[0,1]$
$$
f(x)=\left\{\begin{array}{ll}
x+\frac{1}{2} & 0 \leqslant x \leqslant \frac{1}{2} ; \\
2(1-x) & \frac{1}{2}<x \leqslant 1 .
\end{array}\right.
$$

Try to find five different points $x_{0}, x_{1}, x_{2}, x_{3}, x_{4}$ on $[0,1]$, such that
$$
f\left(x_{0}\right)=x_{1}, f\left(x_{1}\right)=x_{2}, f\left(x_{2}\right)=x_{3}, f\left(x_{3}\right)=x_{4}, f\left(x_{4}\right)=x_{0} .
$$
Notice that $f(0)=\frac{1}{2}, f\left(\frac{1}{2}\right)=1, f(1)=0$, so $f(x)$ has a 3-periodic orbit $\left\{0, \frac{1}{2}, 1\right\}$. By Sarkovskii's theorem, $f(x)$ has periodic orbits of any period, including 5-periodic orbits. The question now is how to find a 5-periodic orbit.

Considering that if $x \in\left[0, \frac{1}{2}\right]$, then $f(x) \in\left[\frac{1}{2}, 1\right]$; if $x \in\left[\frac{1}{2}, 1\right]$, then $f(x)$ can fall on any point in $[0,1]$. Therefore, we can assume that $x_{0} \in\left[0, \frac{1}{2}\right]$, and $x_{1}, x_{2}, x_{3}, x_{4}$ are all in $\left[\frac{1}{2}, 1\right]$. Writing out the $n$-th iteration of $f(x)$ on $\left[\frac{1}{2}, 1\right]$, if $x, f(x), \cdots, f^{[n-1]}(x)$ are in $\left[\frac{1}{2}, 1\right]$, we have
$$
f^{[n]}(x)=(-2)^{n}\left(x-\frac{2}{3}\right)+\frac{2}{3} .
$$

Thus, $x_{0}$ and $x_{1}$ should satisfy the relationship
$$
\left\{\begin{array}{l}
x_{0}+\frac{1}{2}=x_{1}, \\
(-2)^{4}\left(x_{1}-\frac{2}{3}\right)+\frac{2}{3}=x_{0} .
\end{array}\right.
$$

Solving these equations, we get $x_{0}=\frac{2}{15}, x_{1}=\frac{19}{30}$. Then we calculate $x_{2}=f\left(x_{1}\right)=\frac{11}{15}, x_{3}=f\left(x_{2}\right)=\frac{8}{15}$, $x_{4}=\frac{14}{15}$, and it is easy to verify that $f\left(x_{4}\right)=x_{0}$. Therefore, the 5 numbers $\left\{\frac{2}{15}, \frac{19}{30}, \frac{11}{15}, \frac{8}{15}, \frac{14}{15}\right\}$ form the desired 5-periodic orbit.
Note: $x_{0}=\frac{1}{9}$ is also possible.