To find all homomorphisms, it suffices to map a generator of $U(10)$ to some element in $\mathbb{Z}_8$. We can note that 3 generates $U(10)$ so we have 8 options to choose from in $\mathbb{Z}_8$. So there are 8 homomorphisms.
However, the hint on my textbook says that there are only 4. Could someone explain why?
Note that $\mathbb{Z}_8$ has a unique group of order $d$, where $d$ is any divisor of $8$. Indeed any finite cyclic group $G$ has a unique subgroup of order $d$, where $d \mid |G|$. Now obviously the image of the group $U(10) \cong \mathbb{Z}_4$ must be the unique subgroup of order $4,2$ or $1$ of $\mathbb{Z}_8$.
For the first case there are two possibilities, as the subgroup of order $4$ has $\phi(4) = 2$ generators. Those homomorphisms are induced by $3 \to \bar{2}$ and $3 \to \bar{6}$.
For the second case we get that we have a single option, namely the homomorphism induced by $3 \to \bar{4}$.
Finally you have the trivial homomorphism induced by $3 \to \bar{0}$
You can also notice that the image of $\phi: \mathbb{Z}_n \to \mathbb{Z}_m$ is included in the unique subgroup $\mathbb{Z}_{\gcd(m,n)}$ of $\mathbb{Z}_m$. Now $\mathbb{Z}_n$ can be mapped to any subgroup of $\mathbb{Z}_{\gcd(m,n)}$, as every subgroup of $\mathbb{Z}_n$ is normal, so by Correspondence Theorem it is a kernel of such a mapping.
Now subgroups of $\mathbb{Z}_{\gcd(m,n)}$ are of the form $\mathbb{Z}_d$, where $d \mid \gcd(m,n)$. Each of those has exactly $\phi(d)$ generators, so there are $\phi(d)$ homomorphisms to $\mathbb{Z}_d$. So finally the wanted number of homomorphisms is:
$$\sum_{k=1; gcd(d,\gcd(m,n)) \not = 1}^{\gcd(m,n)} \phi(k) = \gcd(m,n)$$
The last identity is well-known one and in general its true that $\sum_{k=1; gcd(d,s) \not = 1}^{s} \phi(k) = s$