, 2020

Solution:

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.