Ask for all solutions (real numbers) of the system of equations
$$
\left\{\begin{array}{l}
x_{2}=\frac{1}{2}\left(x_{1}+\frac{a}{x_{1}}\right), \\
x_{3}=\frac{1}{2}\left(x_{2}+\frac{a}{x_{2}}\right), \\
\cdots \cdots \\
x_{n}=\frac{1}{2}\left(x_{n-1}+\frac{a}{x_{n-1}}\right), \\
x_{1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right)
\end{array}\right.
$$

(1961 Kyiv Mathematical Olympiad)
By the AM-GM inequality, we know $x_{i} \geqslant \sqrt{a}(i=1, \cdots, n)$, then the system of equations has the same solution as the equation $f^{[n]}(x)=x$.

In general, we have $M_{n} \supseteq M_{1}$ and when $f(x)$ is (monotonic increasing) $M_{n}=M_{1}$. For all $x_{1}, x_{2} \in [\sqrt{a},+\infty)$, and $x_{2}>x_{1}$, then
$$
\begin{aligned}
f\left(x_{2}\right)-f\left(x_{1}\right) & =\frac{1}{2}\left[\left(x_{2}-x_{1}\right)+a\left(\frac{1}{x_{2}}-\frac{1}{x_{1}}\right)\right] \\
& =\frac{1}{2}\left[\left(x_{2}-x_{1}\right)-a \frac{x_{2}-x_{1}}{x_{2} x_{1}}\right] \\
& =\frac{1}{2}\left(x_{2}-x_{1}\right)\left(1-\frac{a}{x_{1} x_{2}}\right) \geqslant 0 .
\end{aligned}
$$

Thus, $M_{n}=M_{1}$. Therefore,
$$
f(x)=x \Rightarrow \frac{1}{2}\left(x+\frac{a}{x}\right)=x \Rightarrow x= \pm \sqrt{a} .
$$

Hence, the original system of equations has two solutions $(\sqrt{a}, \sqrt{a}, \cdots, \sqrt{a}),(-\sqrt{a},-\sqrt{a}, \cdots,-\sqrt{a})$.