[Example 4.3.7] Let $n$ be an odd number not less than 3, and $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers. Try to find all invertible functions
$$
f:\left\{x_{1}, x_{2}, \cdots, x_{n}\right\} \rightarrow\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}
$$

such that $\left|f\left(x_{1}\right)-x_{1}\right|=\left|f\left(x_{2}\right)-x_{2}\right|=\cdots=\left|f\left(x_{n}\right)-x_{n}\right|$.
For this problem, we can start from the requirement of the function given in the question, that is, $|f(x_i) - x_i| = a, i = 1, 2, \cdots, n$.
At this point, $f(x_i) = x_i \pm a$.
We study $f(x_i)$ and $x_i$ as a whole.
Notice that $f: \{x_1, x_2, \cdots, x_n\} \rightarrow \{x_1, x_2, \cdots, x_n\}$ is a one-to-one mapping from the set to itself, so overall we have
\[
\sum_{i=1}^{n} f(x_i) = \sum_{i=1}^{n} x_i.
\]

At this point, we have
\[
\begin{aligned}
0 & = \sum_{i=1}^{n} f(x_i) - \sum_{i=1}^{n} x_i = \sum_{i=1}^{n} [f(x_i) - x_i] \\
& = \pm a \pm a \pm a \pm \cdots \pm a = m(a - a) + k a = k a.
\end{aligned}
\]

Since $n$ is odd, $k$ is odd, so from $k a = 0$ and $k$ being odd, we know $a = 0$. Thus, $f(x_i) = x_i$.
Therefore, for all $x \in \{x_1, x_2, \cdots, x_n\}$, we have
\[
f(x) = x.
\]

On the other hand, $f(x) = x$ satisfies the requirements of the problem.
Thus, the problem has a unique solution: $f(x) = x$.