When can a homomorphism be determined entirely by its generators I read a text which says that:
Just because a homomorphism $ϕ :G  →  H$  is determined by the image of its generators does not mean that any such image will work. 
e.g.: Suppose we try to define homomorphism $ϕ :Z_3  →  Z_4$  by $ϕ(1)=1$  , then we get $ϕ(0)=ϕ(1+1+1)=3$  which isn't possible as $ϕ(0)=0$.
Does there exist some case in which a homomorphism is entirely determined by its generators?
 A: By the fundamental theorem on homomorphisms, $\hom(\mathbb{Z}/n\mathbb{Z},G) \cong \{g \in G : g^n=1\}$, the $n$-torsion of $G$. In other words, $g^n=1$ is the only relation which is required by the image $g$ of the canonical generator of $\mathbb{Z}/n\mathbb{Z}$. One can easily derive from this $\hom(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z}/\mathrm{ggT}(n,m)$.
If $E$ generates $G$, then $\hom(G,H) \to \mathrm{Map}(E,H)$ is injective. This is what one means by "a homomorphism is determined by the images of the generators". It is surjective (for all $H$) if and only if $E$ is a free generating set of $G$, i.e. $G= F(E)$ is a free group. Only in this case, every choice of the images produces a homomorphism.
In general, a group presentation exactly contains the information about the relations which are necessary for defining a homomorphism. For example, $G=\langle x,y : x^2 = y^5=1 , xyx^{-1} = y^2 \rangle$ is the group with the property that homomorphisms $G \to H$ correspond to elements $a,b \in H$ (the images of $x,y$) such that $a^2=b^5=1$ and $aba^{-1} = b^2$.
A: I will expand my comment into an answer: 

Theorem: Given a group with presentation 
  $$G = \langle \, g_i \bigm| r_j \, \rangle
$$
  given another group $H$, and given a function $f$ which
  assigns to each $g_i$ a value $f(g_i) \in H$, the function $f$ extends
  to a homomorphism $F:G \to H$ if and only if each $f(r_j)$ is the identity.

Proof: Let $\langle g_i \rangle$ denote the free group with free basis $\{g_i\}$. Let $N$ be the smallest normal subgroup of $\langle g_i \rangle$ containing $\{r_j\}$. By definition of presentation,
$$G = \langle g_i \rangle \, / \, N
$$
Let $q : \langle g_i \rangle \to G$ be the quotient homomorphism.
By the universal property for free groups, the function $f$ extends to a homomorphism $\widetilde F : \langle g_i \rangle \to H$. Let $K = \text{kernel}(\widetilde F)$. Then we have the following chain of equivalences: $\{r_j\} \subset K$ $\iff$ $N < K$ $\iff$ there is a homomorphism $F : G \to H$ such that $F \circ q = \widetilde F$ $\iff$ there is a homomorphism $F : G \to H$ that extends $f$.
A: A homomorphism is always determined by its generators, whether it is an isomorphism or not. To be explicit:

Q: does there exist some case in which a homomorphism is entirely determined by its generators?A: Yes, every single possible case. A homomorphism is always defined by its generators.

All the example is saying is that you cannot just take some map of the generators and hope that it is a homomorphism.
Three more examples:

*

*Any map $\phi: \mathbb{Z}_n\rightarrow\mathbb{Z}$, $1\mapsto x$, for $x\neq 0$, is not a homomorphism as if it is then $n\cdot x=0$, a contradiction!


*Suppose $\gcd(n, m)=1$. Then $\phi: \mathbb{Z}_m\rightarrow\mathbb{Z}_n$, $1\mapsto x$, for $x\neq 0\pmod n$, is not a homomorphism because the image of the subgroup $\mathbb{Z_m}$ must have order dividing $m$ (why?) but all subgroups of $\mathbb{Z}_n$ have order dividing $n$. (This is a generalisation of the example given in your question.)


*Suppose $G$ is simple and $H$ contains no subgroup isomorphic to $G$. Then any map $\phi: G\rightarrow H$ where the generators are not mapped to the identity of $H$ is not a homomorphism. For example, the generators in a map $\phi: A_5\rightarrow \mathbb{Z}_n$ must be sent to the identity, otherwise the map is not a homomorphism.
Of course, you may be asking "when does every choice of generator produce a homomorphism". If that is so, this is not clear. But then you should read Martin's answer!
