When is a functor uniquely determined by where it sends objects? Consider a functor $F : C \to D$ . Let $F_1$ be a map from the objects of $C$ to the objects of $D$ and $F_2$ be a map on the arrows. Under what circumstances is $F$ uniquely determined by $F_1$ ?
Example 1
In a discussion with friends yesterday, the list type constructor [a] in Haskell was brought up as an example of an endofunctor $G$ of "Hask" (the category whose objects are Haskell types and morphisms are values). We conjectured that there's exactly one way to send arrows to arrows with $G$ once you picked where the types were sent. In other words, let $K$ be an endofunctor of $\text{Hask}$ . If $K_1 = G_1$ , then $K = G$ .
Restating Example 1 without referring to programming. Let $W$ be the category defined as follows:
The objects of $W$ are types in a typed lambda calculus with function types $\to $, products $\times$ , sums $+$ , and isorecursive types $\mu \alpha \mathop{.} T$ . The primitive types are $0$ (void) and $1$ (unit). Note that recursive types exist, but parametric polymorphism does not.
The arrows of $W$ are the values in the typed lambda calculus outlined above that are functions. For instance, $ \lambda x. \pi_1(x) $ is an element of homset from $(1 \times 1)$ to $1$ . The values are only considered up to extensional equality. So $\lambda x . x + x$ and $\lambda x . 2 * x$ denote the same arrow.
The list functor $G$ , sends the object $3$ to $\mu \alpha \mathop{.} 1 + 3 \times \alpha$ ... which is a list of values drawn from the type $3$ . I'm deliberately being vague about whether values of recursive types can be infinite or not. I think the argument works out regardless of whether infinite lists are permitted or forbidden.
I think it's the case that the functor laws constrain $G$ so that an arrow $f$ is mapped to an arrow that applies the function $f$ once to every item in the list and then assembles the list in the same order, without duplicating or deleting elements.
Counterexample 2
Let $M$ be a free category with one object $a$ generated by the non-identity arrow $f : a \to a$ . $M$ is isomorphic to $\mathbb{N}$ . The endofunctor $J$, defined as $J_1(x) = x$ and $J_2(x) = 2x$ , trivially maps the objects the same way as the identity functor does, but is not the identity functor.
Is there a neat set of conditions that, when met, force a functor to be uniquely determined by where it sends objects?
 A: This is quite rare and will pretty much never happen for the typical sort of "category of all structures of a certain type" (as opposed to "diagram-like categories" like posets or monoids).  In particular, suppose that $F:C\to D$ is a functor, $c$ is an object of $C$, and $i:F(c)\to F(c)$ is an automorphism in $D$.  Then we can define another functor $F':C\to D$ as follows.  For all objects $a$, $F'(a)=F(a)$.  For a morphism $f:a\to b$ with $a,b\neq c$, $F'(f)=F(f)$.  For a morphism $f:a\to c$ with $a\neq c$, $F'(f)=iF(f)$.  For a morphism $f:c\to b$ with $b\neq c$, $F'(f)=F(f)i^{-1}$.  Finally, for a morphism $f:c\to c$, $F(f)=iF(f)i^{-1}$.
It is easy to check that this $F'$ is indeed a functor.  Intuitively, it is the same functor as $F$, except that we've "relabeled" $F(c)$ by the isomorphism $i$.  Indeed, $F'$ is naturally isomorphic to $F$, by a natural isomorphism that is the identity on all objects except $c$ but is $i$ on $c$.
Since $F'$ is the same on objects as $F$, this will be a counterexample unless $F'$ happens to be equal to $F$.  This will rarely be the case for every possible choice of $c$ and $i$.  For instance, if $F$ is faithful, this implies the automorphism group of every object of $C$ must be abelian (otherwise pick $i$ to be the image of some noncentral automorphism of an object $c$, and $F'$ will not agree with $F$ on automorphisms of $c$).  This should in particular show that your list functor is actually not uniquely determined by where it sends objects.
