I'd like to clear up something that came up in the comments. There are two natural ways to fit the finite cyclic groups together in a diagram. One is to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, m | n$ given by sending $1$ to $1$. This gives a diagram (inverse system) whose limit (inverse limit) is the profinite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$. This diagram also makes sense in the category of unital rings, since they also respect the ring structure, giving the profinite integers the structure of a commutative ring.
This is not the diagram relevant to understanding the circle group. Instead, one needs to take the morphisms $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}, n | m$ given by sending $1$ to $\frac{m}{n}$. This is the diagram relevant to understanding the cyclic groups as subgroups of their colimit (direct limit), which is, as I have said, $\mathbb{Q}/\mathbb{Z}$. And this group, in turn, "compactifies" or "completes" to the circle group in whichever way you prefer.
(These two diagrams are "dual," though, something which I learned recently when I was asked to prove on an exam that $\text{Hom}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \hat{\mathbb{Z}}$. Just observe that $\text{Hom}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Q}/\mathbb{Z}) \simeq \mathbb{Z}/n\mathbb{Z}$ and that contravariant Hom functors send colimits to limits!)