First let us note that $f$ is bijective (on the power set of $\{1,2,\dots,n\}$), i.e. we can reconstruct $S$ from knowing $f(S)$. Indeed, we can first reconstruct whether $1$ is in $S$, then which prime numbers are in $S$, then which products of two primes are in $S$ etc. since each time we already know it for all proper divisors and then by parity also for the new number itself. This proves the claim. OK, so now we are really asked to study the length of the orbit of $\{1\}$ under $f$. Clearly, the first iterate is $\{1,2,\dots,n\}$. The next image is all numbers with an odd number of divisors, i.e. all perfect squares. The next image is all numbers with an odd number of square divisors, i.e. all numbers with all exponents in the prime factorization being $0,1 \pmod{4}$. The next image is then all fourth powers. Iterating this argument, we see that we always get some congruence condition on the exponents, namely $0 \pmod{1} \mapsto 0 \pmod{2}\mapsto 0,1 \pmod{4} \mapsto 0 \pmod{4} \mapsto 0,1,2,3 \pmod{8} \mapsto 0,2 \pmod{8} \mapsto 0,1 \pmod{8} \mapsto 0 \pmod{8} \mapsto \dots$. Here it is clear that the pattern will be $0 \pmod{2^k}$ after $2^k$ steps (easily proved by induction). So the length of the orbit will be $2^k$ if $2^k$ is maximal with $2^{2^k} \le n$ (so that there are $2^k$-th powers besides $1$). Another way to write it would be as $2^{\lfloor \log_2(\log_2(n))\rfloor}$.