Solution:
Define an oriented graph $G$ with vertices the elements of $A$ and oriented edge $xy$ if $f(x)=y$. We have to count the graphs $G$ such that:
- there are no cycles with length greater than $1$;
- there is a chain $a_{2} \ldots a_{n}$ with length $n-2$ and there is no chain with length $n-1$;
- the only edge outside this chain has the form $a_{1} a_{j}$, where $3 \leq j \leq n$;
- there is a unique loop $a_{n} a_{n}$.
The chain can be chosen in $n!$ ways, and the edge outside it - in $n-2$ ways. Note that the graphs for which this edge is $a_{1} a_{3}$, are counted two times and hence their number is equal to $\binom{n}{2}(n-2)!$. So the answer of the problem is
$$n!(n-2)-\binom{n}{2}(n-2)!=\frac{n!(2n-5)}{2}$$