This is the group of units mod n (i.e. the group of units in the ring Z/nZ). Its isomorphism type is fully known. First note that if ##n=p_1^{a_1} \cdots p_r^{a_r}## is the prime factorization of n then $$U(n) = U(p_1^{a_1}) \times \cdots \times U(p_r^{a_r})$$ by the Chinese remainder theorem. Thus it suffices to determine what ##U(p^a)## looks like for a prime p. Here there are two cases to consider.
1) If p is odd then ##U(p^a)## is cyclic (of order ##\varphi(p^a)##).
2) If p=2 then ##U(2)## is trivial (hence cyclic) but for a>1 ##U(2^a) = \mathbb Z / 2\mathbb Z \times \mathbb Z/2^{a-2} \mathbb Z## (which is noncyclic if a>2).
In particular, it follows that U(n) is cyclic iff n=2, 4, p^a or 2p^a, where p is an odd prime.
Depending on how much group theory you know, these results might not be so easy to prove. They follow most easily from the structure theorem for finitely generated abelian groups, which is a nontrivial theorem.
When ##U(n)## is cyclic, a generator is called a primitive root mod n. In general finding a primitive root explicitly isn't an easy task.