I have two categories $C$ and $D$ and a functor $F: C\rightarrow D$. I select an object $A\in ob(C)$, is there guaranteed to be a homomorphism $Aut(A)\rightarrow Aut(F(A))$? It seems like there should be, but I am having a hard time formalizing the argument, given I'm somewhat new to the subject.
If such a homomorphism is guaranteed to exist, do the properties of the functor imply anything about the group homomorphism? Perhaps faithfulness implies injectivity, or fullness implies surjectivity?
I suspect that if the functor isn't injective on the objects that might complicate things.
Everything you've said is correct except fullness doesn't imply surjectivity. Part of the data of a functor is a family of maps $\operatorname{Hom}(X, Y) \to \operatorname{Hom}(F(X), F(Y))$, and in particular we have such a map when $X = Y$. In this case the laws for being a functor say the map $\operatorname{End}(X) \to \operatorname{End}(F(X))$ is a monoid homomorphism, ie preserves the operation (composition) and the identity element (the identity morphism). Any homomorphism between monoids sends invertible elements to invertible elements (exercise) and so restricts to a homomorphism between their maximal subgroups/groups of units. In particular we get a group homomorphism $\operatorname{Aut}(X) \to \operatorname{Aut}(F(X))$.
Fullness implies $\operatorname{End}(X) \to \operatorname{End}(F(X))$ will be surjective, but this property can break when we restrict to groups of units. For example we have a surjective monoid homomorphism $\mathbb{N} \times \mathbb{N} \to \mathbb{Z}$ given by $(n, m) \mapsto n - m$, but the only invertible element of $\mathbb{N} \times \mathbb{N}$ is $(0, 0)$ while everything in $\mathbb{Z}$ is invertible. We can define one object categories where the endomorphisms of the objects are $\mathbb{N} \times \mathbb{N}$ and $\mathbb{Z}$, and the homomorphism above will become a functor between them which is full but not surjective on automorphism groups.