$\mathbb Z/n\mathbb Z$ is a group under addition modulo $n$; multiplication modulo $n$ is well defined, but does not make $\mathbb Z/n\mathbb Z$ a group for $n>1$ since $0$ does not have a multiplicative inverse.
Recall for an abelian group $A$, one denotes the operation by $+$, the identity element by $0$, and for $a \in A$ and $n \in \mathbb N$, $a^n$ becomes $na = a + a +\cdots + a$ $n$-times.
Say $f: \mathbb Z/4\mathbb Z \to \mathbb Z/6\mathbb Z$ is a group homomorphism. Since $f$ is a group homomorphism, $f(0) = 0$. Now since $\mathbb Z/4\mathbb Z$ is a cyclic group generated by $1 \pmod 4$, $f$ is completely determined by $f(1)$. Since $4 f(1) = f(1) + f(1) + f(1) + f(1) = f(1+ 1 + 1 +1) = f(4) = f(0) = 0$, we see that the (additive) order of $f(1)$ divides $4$. There is no element of order $4$ in $\mathbb Z/6\mathbb Z$, so the order of $f(1)$ can be $1$ or $2$. If it's 1, then $f(1) = 0$ and $f$ is the zero map; if it's 2, then $f(1) =3$.
So there are two homomorphisms.