A group $G$ can be thought of as a groupoid $G$ with a single object. How is defining an action of $G$ on an object of a category $C$ the same thing as defining a functor $G \rightarrow C$.
I know that this is a covariant functor but how is it the same?
$G$ is regarded as a category $(*,\text{Hom}(*,*))$, where $*$ is one object, and the set of morphisms from $*\to *$ is given by the elements of $G$ with composition taking place in the obvious way. A group action is equivalent to a group homomorphism $G\to \text{Aut}(X)$, where $X$ is an object of the category $\mathcal{C}$. Now, a functor $F:G\to \mathcal{C}$ is tantamount to choosing an object $X=F(*)$, and a map $$\text{Hom}(G,G)\to\text{Hom}(X,X).$$ Now, because all of the elements of $\text{Hom}(G,G)$ are isomorphisms, $\text{Hom}(G,G)=\text{Aut}(G)$. By the functoriality property of $F$ applied to morphisms, we know that isomorphisms are sent to isomorphisms, so actually our map is $\text{Hom}(G,G)=\text{Aut}(G)\to \text{Aut}(X)$. Finally, by the functoriality properties, $F(1_G)=1_X$ and $F(\phi\circ \psi)=F(\phi)\circ F(\psi)$. It follows that $\text{Aut}(G)\to \text{Aut}(X)$ is a group homomorphism, and we have defined a group action of $G$ on $\mathcal{C}$.