Question 20 Let $f(x)$ be a function on $[0,1]$
$$
f(x)=\left\{\begin{array}{ll}
2 x+\frac{1}{3} & 0 \leqslant x<\frac{1}{3} ; \\
\frac{3}{2}(1-x) & \frac{1}{3} \leqslant x \leqslant 1 .
\end{array}\right.
$$

Try to find 5 different points $x_{0}, x_{1}, x_{2}, x_{3}, x_{4}$ on $[0,1]$ such that
$$
f\left(x_{0}\right)=x_{1}, f\left(x_{1}\right)=x_{2}, f\left(x_{2}\right)=x_{3}, f\left(x_{3}\right)=x_{4}, f\left(x_{4}\right)=x_{0} .
$$
If $x \in\left[0, \frac{1}{3}\right)$, then $f(x) \in\left[\frac{1}{3}, 1\right)$; if $x \in\left[\frac{1}{3}, 1\right]$, then $f(x) \in[0,1]$.

Assume $x_{0} \in\left[0, \frac{1}{3}\right), x_{1}, x_{2}, x_{3}, x_{4} \in\left[\frac{1}{3}, 1\right]$, it is easy to find the iterative expression of $f(x)$ on $\left[\frac{1}{3}, 1\right)$:
$$
f^{[n]}(x)=\left(-\frac{3}{2}\right)^{n}\left(x-\frac{3}{5}\right)+\frac{3}{5} .
$$

Thus, $x_{0}$ and $x_{1}$ satisfy the relationship:
$$
\begin{array}{l}
\left\{\begin{array}{l}
2 x_{0}+\frac{1}{3}=x_{1}, \\
\left(-\frac{3}{2}\right)^{4}\left(x_{1}-\frac{3}{5}\right)+\frac{3}{5}=x_{0} .
\end{array}\right. \\
\Rightarrow\left\{\begin{array} { l } 
{ 6 x _ { 0 } - 3 x _ { 1 } + 1 = 0 , } \\
{ 1 6 x _ { 0 } - 8 1 x _ { 1 } + 3 9 = 0 . }
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{0}=\frac{6}{73}, \\
x_{1}=\frac{109}{219} .
\end{array}\right.\right.
\end{array}
$$

Substituting in, we get:
$$
\begin{array}{l}
x_{2}=\frac{3}{2}\left(1-\frac{109}{219}\right)=\frac{55}{73}, \\
x_{3}=\frac{3}{2}\left(1-\frac{55}{73}\right)=\frac{27}{73}, \\
x_{4}=\frac{3}{2}\left(1-\frac{27}{73}\right)=\frac{69}{73}, \\
x_{5}=\frac{3}{2}\left(1-\frac{69}{73}\right)=\frac{6}{73} .
\end{array}
$$