Example 5 Let the set $A=\{1,2,3,4,5,6\}$, and the mapping $f: A \rightarrow A$, such that its third composition $f \cdot f \cdot f$ is the identity mapping. How many such $f$ are there?
(1996 Japan Mathematical Olympiad Preliminary)
From the above Theorem 2 and Theorem 1, we know that $f$ is a bijection. In $A$, there only exist mapping cycles of $f$, and each mapping cycle contains 1 or 3 elements. Therefore, $f$ has only the following 3 cases:
(1) $f$ is the identity mapping, and there is only 1 such $f$.
(2) 3 elements in $A$ form a mapping cycle of $f$, and the other 3 elements are fixed points of $f$. There are $C_{6}^{3} \cdot (3-1)!=40$ such $f$.
(3) The elements in $A$ are evenly divided into two groups, each containing 3 elements that form a mapping cycle of $f$. There are $\frac{1}{2} C_{6}^{3} C_{3}^{3} (3-1)! (3-1)!=40$ such $f$.
Therefore, the total number of $f$ is $1+40+40=81$.