Show that there is no other isomorphim Recall: Let $(X,\leq)$, $(Y,\leq ')$ be two partially ordered sets. Let $f:X\rightarrow Y$ be a function such that 
$a\leq b \rightarrow f(a)\leq 'f(b)$
we say $f$ preserves order relation.
If $f$ is a bijection and $a\leq b \iff f(a)\leq 'f(b)$ fir any $a,b\in X$ we say $f$ is an order isomorphism.


*

*Let us define $\leq$ on $\mathbb{N}$ as follows:


$x\leq y \iff$ $x$ divides $y$
Find all order preserving isomorphism from $(\mathbb{N},\leq)$  to $(\mathbb{N},\leq)$.
Since $1$ divides $1$, $2$ divides $2$, ..., then $1\leq 1$, $2\leq 2$,...
, that is $Id_{\mathbb{N}}$.
If $n$ is a natural number, let $n_0^{m_0}\cdots n_{k-1}^{m_{k-1}}$ be the prime factorization.
Let $g:\Bbb{P\to P}$ a bijective function(where $\Bbb P$ is the set of primes).
Then $f(n)=f\left(n_0^{m_0}\cdots n_{k-1}^{m_{k-1}}\right)=g\left(n_0\right)^{m_0}\cdots g\left(n_{k-1}\right)^{m_{k-1}}$.
My question is: How can I show that there is no other isomorphism?
 A: Let $g$ be an automorphism. Then for each prime $p$ we have $1 < p$ and thus $1 < g(p)$. If $g(p)$ were not prime, we'd have some element $1 < x < g(p)$ and thus $1 < g^{-1}(x) < p$ which is absurd. Hence each prime $p$ is sent to another prime $s(p)$. Now, we claim that
$$
g(p_1^{n_1} \cdots p_k^{n_k}) = s(p_1)^{n_1} \cdots s(p_k)^{n_k} \tag{1}
$$
and since these are in fact automorphisms, this characterizes them. In effect, we can do induction on the maximum exponent of a prime decomposition of $n \in \mathbb{N}$. If $n = 1$ or $n$ prime, we have already proved this. Now take $n = p_1^{n_1} \cdots p_k^{n_k} \in \mathbb{N}$ and without loss of generality, let's assume $n_1 \leq \dots \leq n_k$. Since $ p_1^{n_1} \cdots p_k^{n_k-1} <  p_1^{n_1} \cdots p_k^{n_k}$, then
$$
 s(p_1)^{n_1} \cdots s(p_k)^{(n_k-1)} = g(p_1^{n_1} \cdots p_k^{(n_k-1)}) <  g(p_1^{n_1} \cdots p_k^{n_k})
$$
and so $g(p_1^{n_1} \cdots p_k^{n_k}) = s(p_1)^{m_1} \cdots s(p_k)^{m_k}$ with $m_i \geq n_i$. We ought to see that $m_i = n_i$ for all $i$. If for some $j$ it would be $m_j > n_j$, then we would have that
$$
s(p_1)^{n_1} \cdots s(p_k)^{(n_k-1)} < s(p_1)^{n_1} \cdots s(p_k)^{n_k} < s(p_1)^{m_k} \cdots s(p_k)^{m_k}
$$
and therefore applying $g^{-1}$ we get that
$$
p_1^{n_1} \cdots p_k^{(n_k-1)} < g^{-1}(s(p_1)^{n_1} \cdots s(p_k)^{n_k}) < p_1^{n_1} \cdots p_k^{n_k}.
$$
which is absurd. This concludes the proof of $(1)$ and so autmorphisms correspond to bijections of $\mathbb{P}$, and via their enumeration, the latter correspond to bijections of $\mathbb{N}$. In other words, if $p_n$ in the $n$-th prime and we define
$$
\begin{align}
\Gamma : \operatorname{Aut}(&\mathbb{N}, |) \longrightarrow \mathbb{S}(\mathbb{P}) \longrightarrow \mathbb{S}(\mathbb{N})\\
& g \longmapsto  (p \mapsto g(p)) \mapsto s_g 
\end{align}
$$
with $s_g(m) = n$ iff $g(p_m) = p_n$, then $\Gamma$ is bijective.
Edit: according to comments the question concerns automorphisms of the order induced by division, hence what follows does not apply.
To ease on the notation, I will use $U = (\mathbb{N},\leq)$ for the naturals with the usual order and $D = (\mathbb{N},\leq')$ for the naturals with the order given by division. Note that any function $g : U \to D$ which verifies $x \leq y \Rightarrow g(x) \leq' g(y)$ is monotone as a function $\tilde{g}: U \to U$, because $x | y$ implies $x \leq y$. Thus if $g$ were an isomorphism, we would have a bijective monotone function $\tilde{g} : U \to U$ with $g(n) = \tilde{g}(n)$. As we have ruled out in the comments, as a function we have $g \neq id$ and so by well ordering of the naturals (with respect to the order on $U$), 
$$
l := \min \{n : g(n) \neq n\}
$$ 
is well defined, and so $g(l) \neq l$. By surjectivity, $l = g(k)$ and thus $k \neq l$. By minimality, since $k \neq g(k)$ necessarily $l < k$ and so $g(l) < g(k) = l$. By minimality once again, this says that $g(g(l)) = g(l) = g(g(k))$ and so by injectivity, we get that $g(l) = g(k) = l$, and that is absurd. Since the contradiction came from assuming that such $g$ existed, no isomorphism between $U$ and $D$ exists. 
A: The key insight is that for the order you defined, a prime number $p$ has the property that from $1 \le x \le p$ follows $x=1$ or $x=p$. In addition, no other number (besides 1, which is an obvious fixpoint of any isomorphism) fullfills that condition. So by necessitiy, any order isomorphism must bijectively map primes to primes.
