Is U(1) isomorphic to interet mod n?
I assume that $\displaystyle U(1)$ is the unitary group of one by one matrices? Clearly (not to insult your intelligence) $\displaystyle \phi:U(1)\to\mathbb{T}:[z]\mapsto z$ is an isomorphism (where $\displaystyle \mathbb{T}$ is the circle group) but it's also clear that $\displaystyle \mathbb{T}\cong\mathbb{R}/\mathbb{Z}$ since the map $\displaystyle \phi:\mathbb{R}\to\mathbb{T}:r\mapsto e^{2\pi i r}$ is a homomorphism with kernel $\displaystyle \mathbb{Z}$ (just use the FIT). This is sort of close to what you want? It's also true that $\displaystyle U(1)\cong\text{SO}(2)$ (the special orthogonal group) where I'm sure it's clear to you that you just send $\displaystyle \mathbb{T}\ni a+bi\mapsto \begin{pmatrix}a & b\\ -b &a\end{pmatrix}$.
So, the answer to your question is no. Is there a particular reason you think that? You know that $\displaystyle \#(U(1))\geqslant\infty$ since $\displaystyle [e^{ir}]$ is an element for $\displaystyle r\in\mathbb{R}$