To prove something is a functor isn't it enough to prove that it commutes with composition? The second thing you usually have to prove is that $F(\text{id}_X) = \text{id}_{FX}$ for all $X \in C$, where $F: C \to C'$ is the supposed functor.
I think it's enough to just prove that $F$ commutes with composition, or $F(g\circ f) = F(g) \circ F(f), \ \forall f: X\to Y, g : Y \to Z$ in $C$.
Proof: if $F$ commutes with composition in $C$, then for all $X \in C, \ f : X \to Y \in \text{Mor}(C)$, we have $F(f) = F(f \circ \text{id}_X) = F(f) \circ F(\text{id}_X)$, but the identity in $\text{Hom}_{C'}(FX, FX)$ is unique, so $F(\text{id}_X) = \text{id}_{FX}$ necessarily.
Is that correct?
 A: Here's a concrete counter-example.  Every monoid (set with a associative binary operation with left and right unit, i.e. a group without inverses) gives rise to a category: pick something arbitrary for the single object and have the hom-set be the elements of the monoid, composition be the multiplication, and the identity be the unit of the monoid.  A functor between such categories is then a monoid homomorphism. Your question is then: is a semigroup homomorphism automatically a monoid homomorphism?  A semigroup being like a monoid but without a unit.
The answer is "no". The powerset of a singleton set, $\mathcal{P}(1)$, is a monoid with binary operation union and unit $\{\}$.   Let $f : \mathcal{P}(1) \to \mathcal{P}(1)$ defined by $f(s) = 1$. Then $1 = f(s\cup t) = f(s)\cup f(t) = 1 \cup 1 = 1$, so it is a semigroup homomorphism, but $f(\{\}) = 1 \neq \{\}$, so it is not a monoid homomorphism.
That said, every surjective semigroup homomorphism is a monoid homomorphism.  This proves that every full functor automatically preserves identities if it preserves composition.
