Example 4 Let $D=\{1,2, \cdots, 10\}, f(x)$ be a one-to-one mapping from $D$ to $D$, and let $f_{n+1}(x)=$ $f\left(f_{n}(x)\right), n \in \mathbf{N}_{+} ; f_{1}(x)=f(x)$. Try to find a permutation $x_{1}, x_{2}, \cdots, x_{10}$ of $D$, such that $\sum_{i=1}^{10} x_{i} f_{2520}(i)=220$.
For any $i(i=1,2, \cdots, 10), f_{0}(i)=i, f_{1}(i), f_{2}(i), \cdots, f_{10}(i)$, at least two of them are equal. Therefore, from $f$ being a one-to-one mapping, it follows that there exists $1 \leqslant r_{i} \leqslant 10$, such that $f_{r_{i}}(i)=i$. Since $2520=2^{3} \times 3^{2} \times 5 \times 7$ is the least common multiple of $1, 2, 3, \cdots, 10$, $r_{i}$ divides $2520 (i=1,2, \cdots, 10)$, which leads to $f_{2520}(i)=i (i=1,2, \cdots, 10)$. According to the sorting principle, $\sum_{i=1}^{10} x_{i} \cdot i \geqslant 10 \times 1 + 9 \times 2 + 8 \times 3 + \cdots + 1 \times 10 = 220$, and the equality holds if and only if $x_{1}, x_{2}, \cdots, x_{9}, x_{10}$ are $10, 9, 8, \cdots, 2, 1$ respectively. Therefore, the required permutation is $10,9,8,7,6,5,4,3,2,1$.