Let $n\geqslant 0$. Suppose that $U$ is a unitary matrix in $M_n$ and there are two unital ${}^\ast$-homomorhpisms $\pi_1\colon M_n\to A, \pi_2\colon M_n \to B$, where $A,B$ are C*-algebras such that
$\pi_1(U)\in A_+$ and $\pi_2(U)\in -B_+$.
Does it follow that $U=0$, and consequently, $n=0$?
Yes it does. As $M_n$ is simple, both homorphisms are monomorphisms. If we replace $A$, $B$ with $\pi_1(M_n)$, $\pi_2(M_n)$ respectively, we may assume that $\pi_1$ and $\pi_2$ are isomorphisms.
In particular, from $\pi_1(U)\in A_+$ we get that $\sigma(U)=\sigma(\pi_1(U))\subset[0,\infty)$. But then $\sigma(U)=\{1\}$, and $U=I$.
With a similar reasoning for $\pi_2$, we get that $U=-I$. So no such $U$ can exist (note that a unitary cannot be zero).