Consider the natural numbers $ \mathbb{N} $ with the order given by the divisibility relation, ``$ \mid $''. An automorphism of $ (\mathbb{N}, \ \mid) $ is a bijective function $ f: \mathbb{N} \rightarrow \mathbb{N} $ verifying $ n \mid m \longleftrightarrow f(n) \mid f(m) $. It is well known that the automorphism of $ (\mathbb{N}, \ \mid) $ can be described as: $$ f(n) = \prod g(p_i)^{k_i} , \ \ \mbox{ with } \ \ n = \prod p_i^{k_i} = \mbox{ prime factorization of } n $$ being $ g : \lbrace \mbox{prime numbers} \rbrace \rightarrow \lbrace \mbox{prime numbers} \rbrace $ a bijection (the product $ \prod p_i^{k_i} $ is extended over all prime numbers, but only a finite number of the exponents $ k_i $ are nonzero).
My question is, there is a similar result for functions that ONLY preserves order, this is $ f: \mathbb{N} \rightarrow \mathbb{N} $ with $ n \mid m \longrightarrow f(n) \mid f(m) $, but not necessarily $ f(n) \mid f(m) \longrightarrow n \mid m $ and not necessarily biyective?