Let $n$ be a positive integer. Suppose for any $\mathcal{S} \subseteq \{1, 2, \cdots, n\}$, $f(\mathcal{S})$ is the set containing all positive integers at most $n$ that have an odd number of factors in $\mathcal{S}$. How many subsets of $\{1, 2, \cdots, n\}$ can be turned into $\{1\}$ after finitely many (possibly zero) applications of $f$?
1. **Understanding the function \( f \)**:
   - The function \( f(\mathcal{S}) \) takes a subset \(\mathcal{S}\) of \(\{1, 2, \cdots, n\}\) and returns a set containing all positive integers at most \( n \) that have an odd number of factors in \(\mathcal{S}\).
   - This function is bijective on the power set of \(\{1, 2, \cdots, n\}\). This means that for any subset \(\mathcal{S}\), we can uniquely determine \(\mathcal{S}\) from \( f(\mathcal{S}) \).

2. **Analyzing the orbit of \(\{1\}\) under \( f \)**:
   - We start with the set \(\{1\}\).
   - Applying \( f \) to \(\{1\}\), we get \(\{1, 2, \cdots, n\}\) because every number has an odd number of factors in \(\{1\}\) (since 1 is a factor of every number).
   - The next application of \( f \) gives us the set of all numbers with an odd number of divisors, which are the perfect squares.
   - The next application of \( f \) gives us the set of all numbers with an odd number of square divisors, which are numbers where all exponents in their prime factorization are \(0\) or \(1 \pmod{4}\).
   - Continuing this process, we see that the sets we obtain are characterized by congruence conditions on the exponents in the prime factorizations of the numbers.

3. **Pattern of the sets**:
   - The pattern of the sets obtained can be described as follows:
     - \(0 \pmod{1} \rightarrow 0 \pmod{2} \rightarrow 0,1 \pmod{4} \rightarrow 0 \pmod{4} \rightarrow 0,1,2,3 \pmod{8} \rightarrow 0,2 \pmod{8} \rightarrow 0,1 \pmod{8} \rightarrow 0 \pmod{8} \rightarrow \dots\)
   - This pattern shows that after \(2^k\) steps, we get the set of numbers that are \(0 \pmod{2^k}\).

4. **Length of the orbit**:
   - The length of the orbit of \(\{1\}\) under \( f \) is \(2^k\) where \(2^k\) is the largest power of 2 such that \(2^{2^k} \le n\).
   - This can be written as \(2^{\lfloor \log_2(\log_2(n))\rfloor}\).

The final answer is \( \boxed{ 2^{\lfloor \log_2(\log_2(n))\rfloor} } \).