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 $mathbb{Z}/nmathbb{Z} to mathbb{Z}/mmathbb{Z}, m | n$ 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 $mathbb{Z}/nmathbb{Z} to mathbb{Z}/mmathbb{Z}, n | m$ 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 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 $text{Hom}(mathbb{Z}/nmathbb{Z}, mathbb{Q}/mathbb{Z}) simeq mathbb{Z}/nmathbb{Z}$ and that contravariant Hom functors send colimits to limits!)
Edit: Let me also say something about the precise meaning of "compactification" here. A compactification of a space $T$ is an embedding $T to X$ into a compact Hausdorff space $X$ with dense image. The embedding being considered here is the obvious one from $mathbb{Q}/mathbb{Z}$ to $mathbb{R}/mathbb{Z}$, and the fact that it has dense image is essentially what the word "completion" also means. Compactifications are not unique, but it's possible that there is a sense in which as a topological group $mathbb{R}/mathbb{Z}$ is the "most natural" compactification of $mathbb{Q}/mathbb{Z}$. But I don't know too much about topological groups.
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