My question: Let $\cal{C}$ and $\cal{D}$ be two groupoids, i.e. the category whose morphisms are isomorphisms. Let $F:\cal{C}\rightarrow\cal{D}$ be a fully faithful functor from $\cal{C}$ to $\cal{D}$. Then is $F$ injective on objects? In other words, is the object function $F:{\rm{Ob}}(\cal{C})\rightarrow{\rm{Ob}}(\cal{D})$ injective?

Answer: No. Given any set $X$ we can construct a groupoid called the indiscrete groupoid on $X$, which has a unique (iso)morphism $x \to y$ for $x, y \in X$. Every map of sets $f : X \to Y$ induces a fully faithful functor between the corresponding indiscrete groupoids, whether or not $f$ is injective.

What is true is that a fully faithful functor $F : C \to D$ between categories (not necessarily groupoids) induces an injection on isomorphism classes of objects. To see, this let $i : F(c_1) \cong F(c_2)$ be an isomorphism. Because $F$ is full, $i = F(i')$ for some $i' : c_1 \to c_2$, and similarly $i^{-1} = F(j)$ for some $j : c_2 \to c_1$. We have $F(i' \circ j) = F(i') \circ F(j) = \text{id}_{F(c_2)} = F(\text{id}_{c_2})$ so, since $F$ is faithful, $i' \circ j = \text{id}_{c_2}$, and similarly for the other composite. So $j = (i')^{-1}$ and $i' : c_1 \cong c_2$ is an isomorphism. (In the indiscrete groupoid every object has been made isomorphic.)