Example 7 Let $D=\{1,2,3, \cdots, 10\}, f(x)$ be a one-to-one mapping from $D$ to $D$. Let $f_{n+1}(x)=f\left[f_{n}(x)\right]$, $n \in N ; f_{1}(x)=f(x)$. Try to find a permutation $x_{1}, x_{2}, x_{3}, \cdots, x_{10}$ of $D$, such that the following equation holds:
$$
\sum_{i=1}^{10} x_{i} f_{2520}(i)=220 \text {. }
$$
$(1996$, Jiangsu Province High School Mathematics Competition)
Solution: Since $f(x)$ is a one-to-one mapping from $D$ to $D$, for $i=1,2, \cdots, 10$, by $\left\{i, f_{1}(i), f_{2}(i), \cdots\right.$, $\left.f_{10}(i)\right\} \subseteq D$, according to the pigeonhole principle, there exists $r_{i}\left(1 \leqslant r_{i} \leqslant\right.$ $10)$, such that $f_{n_{i}}(i)=i$.

It is easy to see that $2520=2^{3} \cdot 3^{2} \cdot 5 \cdot 7$ is the least common multiple of $1,2, \cdots, 10$, so the above $r_{i} \mid 2520$. Therefore, $f_{2520}(i)$ $=i$ holds for all $i \in D$. Hence, the original equation is
$$
\sum_{i=1}^{10} x_{i} \cdot i=220 .
$$

According to the rearrangement inequality, we have
$$
\begin{aligned}
\sum_{i=1}^{10} i \cdot x_{i} & \geqslant 1 \times 10+2 \times 9+\cdots+10 \times 1 \\
& =220 .
\end{aligned}
$$

Equality holds if and only if $x_{1}, x_{2}, \cdots, x_{10}$ is the reverse permutation of $1,2, \cdots, 10$. Therefore, the required permutation is 10, $9,8, \cdots, 1$.