Example 4 Let $A=\{1,2, \cdots, 10\}$, and the mapping $f: A \rightarrow A$ satisfies $f(x) \neq x(x \in A)$, but $f^{21}(x)=x(x \in A)$. Find the number of such permutations.
【Analysis】For any $i \in A$, we have
$$
i \xrightarrow{f} f(i) \xrightarrow{f} f^{2}(i) \xrightarrow{f} \cdots \xrightarrow{f} f^{j}(i) \xrightarrow{f} \cdots
$$

Thus, for any $j \in \mathbf{N}$, we have
$$
f^{j}(i) \in A,
$$

where $f^{0}(i)=i, f^{j}(i)=f\left(f^{j-1}(i)\right)\left(j \in \mathbf{N}_{+}\right)$.
Since $|A|=10$, for any $j>10$, there must exist $k \in\{1,2, \cdots, 10\}$, such that
$$
f^{j}(i)=f^{k}(i).
$$

Therefore, there exists a smallest $t_{i} \in\{2,3, \cdots, 10\}$, such that $f^{t_{i}}(i)=i$.
This forms a "mapping cycle" with $t_{i}$ elements:
$$
i \rightarrow f(i) \rightarrow \cdots \rightarrow f^{t_{i}-1}(i) \rightarrow f^{t_{i}}(i)=i.
$$

According to the problem, the mapping $f: A \rightarrow A$ consists of one mapping cycle with three elements and one mapping cycle with seven elements, each cycle being a circular permutation. Therefore, the number of mappings that satisfy the given conditions is
$$
2!\times 6!C_{10}^{3}=172800
$$