I'm grappling with the question of viewing a group as a category, which as I understand it means that if $(G,+)$ is the group in question, then the group can also be thought of as a small concrete category $C$ in which ${\rm Ob}(C) =\{ \bullet \}$ and ${\rm Hom}(C,C)$ is just the elements of $G$.
But that's still pretty vague, so I'm trying to construct the specific category $C=(\mathbb{Z_2},+)$, i.e., the integers modulo $2$ under plus, but viewed as a category. What I have so far is that ${\rm Ob}(C) =\{ \bullet \}$ and ${\rm Hom}(C,C)=\{0,1\}$ where $0$ is the identity morphism so that whatever $1$ is (and that part isn't really clear yet), $0 \circ 0 =0 $ and $ 0 \circ 1 = 1 \circ 0 = 1.$ So far, so good. For this category to properly mimic the group, it must be the case that $1 \circ 1=0$.
So can I just define $1$ as a morphism that is an its own inverse? If not, what would be the next step?
Thanks for any advice.
Yes, you can define composition of morphisms as you like, as long as they satisfy the correct axioms.
Notice that in your case ($G=\mathbb Z/2\mathbb Z$) you don't even have to specify this: if you see $G$ as a category with one object (a monoid), all your morphisms will have to be isomorphisms (so that your only hom-set behaves just like the group).
If you have only two morphisms $\{0,1\}$, and $0$ is the identity, what will the inverse of $1$ be?
(Of course, you could also define $1\circ 1$ to be $1$; but as such it wouldn't be an isomorphism, and your monoid wouldn't represent a group anymore.
Conversely, $1\circ0=1=0\circ1$ are forced by the identity axiom.)