Order zero maps in matrix algebra Let $a$ and $b$ are two elements in a $C^*$algebra $A$. We say $a\perp b$ if $ab=ba=a^*b=ab^*=0$. We say a completely positive map $\phi: A \rightarrow B$ is of order zero if for any positive elements $a,b \in A$, $a\perp b \Rightarrow \phi(a) \perp \phi(b)$.
Suppose we consider only matrix algebras (i.e. $\mathcal{B}(\mathbb{C}^n)$ for some $n$). Does there exists, nontrivial order zero cp maps? By nontrivial, I mean not maps like $a \mapsto uau^*$, where $u$ is an unitary (or isometry).

By the structure theorem of Winter-Zacharias (http://arxiv.org/abs/0903.3290) any c.p.c. order zero map $\phi:A\to B$ has the form
$$\phi(a) = h\pi(a) = \pi(a)h,$$
where $\pi:A\to M(C)$ is a $*$-homomorphism from $A$ to the multiplier algebra of the C*-algebra $C$ generated by the image of $\phi$, and $h$ is a positive element in $M(C)$ that commutes with the image of $\pi$. If both $A$ and $B$ are matrix algebras, say $M_n$ and $M_m$, then you have
$$\pi:M_n\to M_m$$
which is either trivial (i.e. zero) or injective, since $M_n$ is simple. In the latter case of course you need $m = kn$, $k\in\mathbb N$ (plus some degeneracy that I'm neglecting here), and $h$ in this case is just $\phi(1_{M_n})$. So loosely speaking, $\pi$ is, up to a unitary conjugation, a $k$-fold embedding of $M_n$ into $M_m$, so that $\phi$ will look like
$$\phi(m) = h^{1/2}u(m\otimes 1_{M_k})u^*h^{1/2},\qquad\forall m\in M_n.$$ 
