If you do modulo arithmetic on a circle you have a certain group. But if you fourier transform a function on a circle you get discrete values. (Compared with continuous fourier tranform of a function on the reals.)
Does this mean the groups of addition on a circle and the group of addition on integers are somehow the same?
In the analysis abelow I consider the circle to be $ \mathbb S^1 = \{z\in\mathbb C: |z|=1\}$. It can be described in a diferent way, but that's the representation I choose.
Adding numbers on a circle can be considered a group of shifts of the circle, yes, that group is usually denoted $U(1)$. It's elements can be denoted as $\rho_a$, $a\in[0,2\pi)$ and it acts on the points of a circle as follows: $$ \rho_a : \mathbb S^1 \ni e^{i\phi} \mapsto e^{i(\phi+a)} \in \mathbb S^1 $$ It can also be represented on functions defined on the circle. for $f : \mathbb S^1 \rightarrow \mathbb C$ one can define $R_af : \mathbb S^1 \rightarrow \mathbb C$ as follows $$ (R_a f)(z) = f(\rho_a^{-1}(z))= f(e^{-ia}z)$$ (The origin of -1 is that to "push" a function in one direction you need to calculate it in an argument "pushed" in the opposite direction.)
The fourier transform of a function defined on a circle is a function defined on integers, but you need to consider how this group acts on the Fourier transform of a function.
If you define $$ \tilde{f_n} = \frac{1}{2\pi i}\oint_{|z|=1} f(z) z^{n-1} dz$$ which gives you the coefficient of the Fourier series when $f$ is defined on an actual circle (and not on a looped interval), then you can find that $$ (\tilde{R_a} \tilde f)_n := (\widetilde{R_a f})_n = e^{-i an} \tilde{f_n}$$ That means that while the group of shifts of the circle is represented on the set of functions defined on $\mathbb Z$ it has nothing to do with the group of additions on integeres, which would be acting on the functions as follows $$ (\tilde{T_a} \tilde f)_n = \tilde{f}_{n-a} $$