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.