For a prime $p$ and positive integer $n$, we let $v_{p}(n)$ denote the largest nonnegative integer $k$ such that $p^{k} \mid n$. Note that $f$ is determined by its action on primes. Since $f(101!)=101$!, by counting prime factors, $f$ must permute the set of prime factors of 101!; moreover, if $p$ and $q$ are prime factors of 101! and $f(p)=q$, we must have $v_{p}(101!)=v_{q}(101!)$. This clearly gives $f(2)=2, f(5)=5$, so it suffices to find the number of possible values for $f(43 \cdot 47 \cdot 101)$. (We can factor $2021=45^{2}-2^{2}=43 \cdot 47$.) There are 4 primes with $v_{p}(101!)=2$ (namely, $37,41,43,47$), so there are 6 possible values for $f(43 \cdot 47)$. Moreover, there are 11 primes with $v_{p}(101!)=1$ (namely, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101). Hence there are 66 possible values altogether.