Solution:

All cycle 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 $1 + \binom{9}{5} \cdot 4! = 3025$.