Example 4 Let the set $A=\{1,2,3,4,5,6\}$, and the mapping $f: A \rightarrow A$, such that its third composite mapping $f \cdot f \cdot f$ is the identity mapping. How many such $f$ are there?
(1996. Japan Mathematical Olympiad Preliminary)
Solution: Since the threefold composite mapping on set A is the identity mapping, it follows from Proposition 1 and Proposition 2 that there are three types of mappings $f$ that meet the conditions:
(1) $f$ is the identity mapping;
(2) There exists a three-element mapping cycle $a \rightarrow b \rightarrow c \rightarrow a (a, b, c$ are distinct), and the other three elements are fixed points;
(3) There exist two three-element mapping cycles $a \rightarrow b \rightarrow c \rightarrow a$ and $a' \rightarrow b' \rightarrow c' \rightarrow a' (a, b, c, a', b', c'$ are distinct).
There is only 1 $f$ of type (1).
For type (2), first choose 3 elements $a, b, c$ from 6 elements, which can be done in $C_{6}^{3}=20$ ways. Then, the arrangement of $a, b, c$ in a circle has (3-1)! $=2$ ways, so there are $20 \times 2=40$ such $f$ (s).

For type (3), first divide 6 elements into two groups, which can be done in $C_{6}^{3} \div 2=10$ ways. Each group can be arranged in a circle in (3-1)!(3-1)! =4 ways, so there are $10 \times 4=40$ such $f$ (s).
In summary, the total number of $f$ is
$$
1+40+40=81 \text { (s). }
$$