All cycles lengths in the permutation must divide 5 , which is a prime number. Either $f(x)=x$ for all $x$, or there exists exactly one permutation cycle of length 5 . In the latter case, there are $\binom{9}{5}$ ways to choose which numbers are in the cycle and 4 ! ways to create the cycle. The answer is thus $1+\binom{9}{5} \cdot 4!=3025$.