I know that Legendre Symbol is unique non-trivial group homomorphism from $U(Z_n)$ to $\{-1,1\}$. Is the same true for Jacobi symbol?
I can prove it for the case when $n$ is power of odd prime. In this case the kernel of non-trivial homomorphism must be subgroup of size $\frac{\phi(n)}{2}$ and $U(Z_n)$ is cyclic thus having one and only one subgroup of this size. So if we have two non-trivial homomorphisms their kernels are equal and so the functions agree on the whole domain since there are only two values they can have.
What about the general case? I tried to play with Chinese Remainder Theorem but got stuck. So any help will be highly appreciated
Write $n=p_1^{a_1}...p_r^{a_r}$ the decomposition of odd $n$ into the products of odd prime numbers. By the Chinese Remainder Theorem :
$$U(Z_n)\text{ is isomorphic to } U(Z_{p_1^{a_1}})\times...\times U(Z_{p_r^{a_r}}) $$
It follows that (one needs to prove something here) :
$$Hom(U(Z_n),\{\pm 1\})=Hom(U(Z_{p_1^{a_1}}),\{\pm 1\})\times ...\times Hom(U(Z_{p_1^{a_1}}),\{\pm 1\})$$
By what you have already done for powers of prime (which works perfectly well) this last set has cardinal $2^r$. Now you must be able to conclude.