Semigroups with no morphisms between them

Let $\mathbb N^+$ be the natural numbers without 0, and consider it as a semigroup under addition. Then there can be no morphism $f: A \to \mathbb N^+$ where $A$ is finite, because then the image $f$ will be bounded by some $n \in \mathbb N^+$. Now trying to add an element $a \in A$ to itself $n+1$ times cannot be respected by the map: $$ f(\underbrace{a + \ldots + a}_{n+1 \text{ times}}) = \underbrace{f(a) + \ldots + f(a)}_{n+1 \text{ times}} \geq n+1, $$ which contradicts that the range of $f$ is bounded by $n$.

No morphism in either direction

Choose two distinct primes $p,q\in\Bbb N_+$ and consider the additive semi-groups

$$\begin{align} P&:=\Big\{\frac n{p^m}\mid n,m\in\Bbb N_+\Big\}\subseteq\Bbb Q_+,\\ Q&:=\Big\{\frac n{q^m}\mid n,m\in\Bbb N_+\Big\}\subseteq\Bbb Q_+. \end{align}$$

Assume there is a morphism $\phi:P\to Q$ and let $a/q^b:=\phi(1)\in Q$ for some $a,b\in\Bbb N_+$. Further, for every $m\in\Bbb N_+$ let

$$\frac{a_m}{q^{b_m}}:=\phi\Big(\frac1{p^m}\Big)\in Q, \qquad\text{for some $a_m,b_m\in\Bbb N_+$}.$$

This means

$$\begin{align} p^m\cdot \frac{a_m}{q^{b_m}}&=\phi\Big(p^m\cdot \frac1{p^m}\Big)\\ &=\phi(1)\\ &=\frac a{q^b}, \end{align}$$

which implies $$p^mq^b\cdot a_m=q^{b_m}\cdot a.$$

Since the left side is divisible by $p^m$, so must be the right side. Since $p$ and $q$ are distinct primes, $a$ must be divisible by $p^m$ for all $m\in\Bbb N_+$, which is a contradiction. Hence there cannot be such a morphism, and since the argument is symmetric, there is no such morphism in either direction.

$\square$

There's no morphism from $\{0\}$ (or from any semigroup with idempotent) to the additive semigroup on $\{2,4,6,\dots\}$