Example 4 Let the set $A=\{1,2, \cdots, 10\},$ and the mapping $f$ from $A$ to $A$ satisfies the following two conditions:
(1) For any $x \in A, f_{30}(x)=x$.
(2) For each $k \in \mathbf{N}_{+}, 1 \leqslant k \leqslant 29$, there exists at least one $a \in A$ such that $f_{k}(a) \neq a$.

Find the number of such mappings $f$ (it is agreed that $f_{1}(x)=f(x), f_{k+1}(x)=f\left(f_{k}(x)\right), k=1,2, \cdots$).
First, we prove that $f$ is a bijection. In fact, for any $x, y \in A, x \neq y$, if $f(x)=f(y)$, then $x=f_{30}(x)=f_{30}(y)=y$, which is a contradiction, so $f$ is injective. Also, for any $y \in A$, let $x=f_{29}(y) \in A$, then $f(x)=f_{30}(y)=y$, so $f$ is surjective, hence $f$ is a bijection.

Next, we prove that in $A$ there only exist mapping cycles of $f$, and no mapping chains. That is, for any $a_{i} \in A$, there exists $k_{i} \in \mathbf{N}_{+}$ such that $f_{k_{i}}\left(a_{i}\right)=a_{i}$, and $f_{s}\left(a_{i}\right) \neq a_{i}\left(s \in \mathbf{N}_{+}, 1 \leqslant s \leqslant k_{i}-1\right)$. In fact, for any $a_{i} \in A$, we have $f\left(a_{i}\right)=a_{i}$ (in this case $k_{i}=1$) or $f\left(a_{i}\right)=a_{i_{1}} \neq a_{i}$. In the latter case, it is impossible to have $f_{2}\left(a_{i}\right)=f\left(a_{i_{1}}\right)=a_{i_{1}}$, otherwise $f\left(a_{i}\right)=f\left(a_{i_{1}}\right)=a_{i_{1}}$ and $a_{i} \neq a_{i_{1}}$, which contradicts the injectivity of $f$, so it must be that $f_{2}\left(a_{i}\right)=f\left(a_{i_{1}}\right)=a_{i}$ (in this case $k_{i}=2$) or $f_{2}\left(a_{i}\right)=a_{i_{2}} \neq a_{i}, a_{i_{1}}$. If $f_{2}\left(a_{i}\right)=a_{i_{2}}$, then by the same reasoning, it is impossible to have $f_{3}\left(a_{i}\right)=f\left(a_{i_{2}}\right)=a_{i_{1}}, a_{i_{2}}$, and it can only be that $f_{3}\left(a_{i}\right)=a_{i}$ (in this case $k_{i}=3$) or $f_{3}\left(a_{i}\right)=a_{i_{3}} \neq a_{i}, a_{i_{1}}, a_{i_{2}}$, and so on. Since $A$ is a finite set, after a finite number of times (let's say $k_{i}$ times), we have $f_{k_{i}}\left(a_{i}\right)=a_{i}$ and $f_{s}\left(a_{i}\right) \neq a_{i}\left(s \in \mathbf{N}_{+}, 1 \leqslant s \leqslant k_{i}-1\right)$. That is, for any $a_{i} \in A$, there exists a mapping cycle containing $k_{i}$ different elements:
$$
a_{i} \rightarrow a_{i_{1}} \rightarrow \cdots \rightarrow a_{i_{i_{i}-1}} \rightarrow a_{i}.
$$

Furthermore, we prove that the number of elements $k_{i}$ in each mapping cycle divides 30. In fact, let $30=q_{i} k_{i}+r_{i}$ $\left(q_{i} \in \mathbf{N}_{+}, r_{i} \in \mathbf{N}, 0 \leqslant r_{i} \leqslant k_{i}-1\right)$. If $r_{i} \neq 0$, then $a_{i}=f_{30}\left(a_{i}\right)=f_{r_{i}}\left(f_{q_{i} k_{i}}\left(a_{i}\right)\right)=f_{r_{i}}\left(a_{i}\right)$, which contradicts $f_{s}\left(a_{i}\right) \neq a_{i}\left(s \in \mathbf{N}_{+}, 1 \leqslant s \leqslant k_{i}-1\right)$, so $r_{i}=0$, i.e., $k_{i} \mid 30$.

From the above analysis, we know that in $A$ there only exist mapping cycles of $f$ and the least common multiple $m$ of the number of elements in each mapping cycle must be a divisor of 30. If $m<30$, then for any $a \in A$, we have $f_{m}(a)=a$, which contradicts the given condition (2), so $m=30$, and the sum of the number of elements in each mapping cycle should equal the number of elements in $A$, which is 10. But $30=5 \times 3 \times 2$ and $10=5+3+2$, so the 10 elements in $A$ exactly form 3 mapping cycles of $f$, with the number of elements in each cycle being 5, 3, and 2, respectively. Therefore, the number of $f$ is equal to the number of ways to choose 5 elements from 10, then 3 from the remaining 5, and finally 2 from the remaining 2, and arrange them in a circular permutation, i.e., the number of $f$ is
$$
C_{10}^{5} \cdot(5-1)!\cdot C_{5}^{3}(3-1)!\cdot C_{2}^{2}(2-1)!=120960
$$