, 2022

$f(S) = (\min S)^\alpha, \forall S \in \mathcal{X}$, where $\alpha \ge 1$. It is easy to see that all such functions are solutions. We now prove that only functions of this form satisfy all the conditions.

Since $\{1\} \cdot \{1\} = \{1\}$, $\mathbb{N} \cdot \mathbb{N} = \mathbb{N}$, we have
$$f(\{1\})^2 = f(\{1\}), \quad f(\mathbb{N\})^2 = f(\mathbb{N\}) \implies f(\{1\}) = f(\mathbb{N\}) = 1. }}}$$

For all sets $S$ containing 1, by condition (i), $1 = f(\mathbb{N\}) \le f(S) \le f(\{1\}) = 1}$, that is,
$f(S) = 1$.

For positive integer $n$, define $g(n) = f(\{n\})$; then condition (ii) tells us
$$g(m) + g(n) \le g(m + n), \quad g(mn) = g(m)g(n).$$
The former inequality implies that $g$ is a strictly increasing function on $\mathbb{N}$. Choose $\alpha$ such that $g(2) = 2^\alpha$; then $2^\alpha = g(2) \ge g(1) + g(1) = 2$ tells us that $\alpha \ge 1$.

We now prove $g(n) = n^\alpha$: for $n = 1, 2$ this is obvious. For any $n \ge 3$ and positive integer $k$, take a positive integer $\ell$ such that $2^\ell \le n^k < 2^{\ell+1}$. Since $g$ is completely multiplicative, we have
$$2^{\ell\alpha} \le g(n)^k < 2^{(\ell+1)\alpha} \implies 2^{\ell/k} \le g(n)^{1/\alpha} < 2^{(\ell+1)/k}.$$
Since $2^{\ell/k} \le n < 2^{(\ell+1)/k}$, we know
$$2^{-1/k} < \frac{2^{\ell/k}}{n} \le \frac{g(n)^{1/\alpha}}{n} < \frac{2^{(\ell+1)/k}}{n} \le 2^{1/k}.$$

Since this holds for all positive integers $k$, $g(n)^{1/\alpha}$ must equal $n$, that is, $f(\{n\}) = g(n) = n^\alpha$.

For any set $S$ not containing 1, let $m = \min S$. Since $\{m\} \subseteq S$, by condition (i) and the above result,
$$f(S) \le f(\{m\}) = m^{\alpha}. \qquad (1)$$
Since $1 = m - (m-1) \in S - \{m-1\} := \{s - (m-1) \mid s \in S\}$, we have
$$f(S) \ge f(\{m-1\}) + f(S - \{m-1\}) = (m-1)^{\alpha} + 1.$$
Replacing $S$ in the above with $S^k := \{s_1 \cdots s_k \mid s_1, \dots, s_k \in S\}$, we obtain
$$f(S)^k = f(S^k) \ge (\min(S^k) - 1)^\alpha + 1 = (m^k - 1)^\alpha + 1.$$
Therefore, when $k > \alpha$,
$$f(S) > (m^k - 1)^{\alpha/k} = m^{\alpha} \left(1 - \frac{1}{m^k}\right)^{\alpha/k} \ge m^{\alpha} \left(1 - \frac{1}{m^k}\right).$$
Since this holds for all positive integers $k > \alpha$, we have $f(S) \ge m^\alpha$. Combining this with (1) gives
$$f(S) = m^{\alpha} = (\min S)^{\alpha}.$$