How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$? I just need a small hint, not the full answer. I know that, if $f$ is a morphism,


*

*$a \mid b \implies f(a) \mid f(b)$

*$a \mid b$ and $a \mid c \implies a \mid b+c$, so $f(a) \mid f(b),   f(a) \mid f(c), f(a) \mid f(b + c)$. Also, $f(a) \mid f(b) + f(c)$

*$\forall a \in \mathbb{N}, a \mid 0$, so $\forall a \in \mathbb{N}, f(a) \mid f(0)$

*$\forall a \in \mathbb{N}, 1 \mid a$, so  $\forall a \in \mathbb{N}, f(1) \mid f(a) $


From these I can draw some conclusions:
a. From 3., $f(0) = a$ ($a \neq 0$), $f(\mathbb{N})$ only contains divisors of $a$.
b. From 4., if $f(1) = a$, then $f(\mathbb{N})$ only contains multiples of $a$.
The most general form I can think of for f is this:
$f(x) = ax^{b} (a, b \in \mathbb{N})$, but I can't seem to go any further than this.
Any help is appreciated.
 A: Your observations a,b are fine so far (if $0\in\mathbb N$ at all, that may depend on your local definitions).
Note that $a|b$ with $a>b$ is only possible with $b=0$.
This allows us to construct uncountably many morphisms $f$ with $f(0)=0$:
Assume $n\in\mathbb N$ and we have selected $a_k$ for $0\le k<n$ such that  $a_0=0$ and $0< r,s\le n$ with $r|s$ implies $a_r|a_s$.
Select $$\tag1a_{n}\in\bigcap_{0<k<n\atop k|n} a_k\mathbb N$$
arbitrarily (which is possible as the set on the right contains at least $0$).
By the choice we guarantee that $k|n$ implies $a_k|a_n$.
Therefore, for any such sequence $(a_n)$, letting $f(n)=a_n$ gives us a morphism.
Note that any morphism with $f(0)=0$ can be obtained this way, i.e. the restriction imposed by $(1)$ is necessary.
For morphisms with $f(0)>0$ you correctly observed that we need $f(n)|f(0)$ for all $n$, esp. $f(n)\ne 0$ for all $n$.
We can do almost the same as above, more precisely select 
$$\tag2 a_{n}\in\{d\in\mathbb N\colon d|f(0)\}\cap\bigcap_{0<k<n\atop k|n} a_k\mathbb N$$
this time observing that the set in $(2)$ is nonempty because it contains $f(0)$. However, at each step there are only finitely many choices. Still, this gives another uncountable lot of morphisms (why?) if $f(0)>1$.
A: It's a big hairy beast.  
At all times, we will assume the prime factorization of $n$ is $n=p_1^{a_1}...p_k^{a_k}$.
For example, define a function first for each prime power, and then you can define $\phi(n)=\phi(p_1^{a_1})...\phi(p_k^{a_k})$. That only gives some basic examples, hardly begins to scratch the surface, but the restriction of $\phi$ at any prime power can be almost anything.  (You can't define it arbitrarily for prime powers - for each $p^a$ you can choose $\phi(p^a)/\phi(p^{a-1})$ arbitrarily.)
One non-trivial example is to define $\phi(n) = p_1p_2...$. In this case your result is always square-free. Indeed, the function returns the largest square-free divisor of $n$.
Another non-trivial example would "forget" the power of $2$ is the prime factorization: $\phi(n)$ is the largest odd factor of $n$.
These first two examples preserve $\gcd$ and $\operatorname{lcm}$, which are the meet and join operator of $(\mathbb N,\mid)$.
Another example would be $\phi(n)=x^{\max(a_1,...a_k)}$, for some fixed $x\in\mathbb N$. This does not preserve $\gcd$, I don't think.
It might be worth just looking at examples that only depend on two primes. So morphisms $\phi$ such that $\phi(2^{a_1}3^{a_2}p_3^{a_3}...p_k^{a_k})=\phi(2^{a_1}3^{a_2})$ for all values. Even that is going to be an interesting mess.
