TITLE: When is this map completely positive?
QUESTION [6 upvotes]: Consider the complex $n$-by-$n$ matrices $M_n$. 
Suppose that $A_i$, for $i=1,\ldots,n^2$, satisfy $\mathrm{Tr}(A_i^*
A_j)=\delta_{ij}$, so that together they form an orthonormal basis for
$M_n$. Define a linear map $T \colon M_n \to M_n \otimes M_n$ by $T(A_i) = A_i \otimes A_i$.

Question: when is $T$ completely positive?

For example, if $A_i$ are the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$, then $T$ is completely positive. In fact, I think these might be the only examples. If $T$ is completely positive, then the following are equivalent to $A_i$ being matrix units as in the above example:

each $A_i$ has rank one;
each positive semidefinite $A_i$ has trace one;
the set $\{0,A_1,\ldots,A_{n^2}\}$ is closed under multiplication;
$T(1)$ is idempotent;
$T^*(1) \leq 1$;
$T$ preserves trace.

These are sufficient conditions, but proving they are sufficient doesn't use  $\mathrm{Tr}(A_i^* A_j)=\delta_{ij}$ at all. Are they necessary?

REPLY [2 votes]: I think I finally have the answer to this question (some 7 years later, and 3 years after first hearing of the problem myself, but never mind). TL;DR: the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$ are indeed the only examples. A detailed proof (with diagrams!) will appear shortly on the arXiv, and I will link to it once it's ready, but I will summarise the gist here for posterity. 
A linear map $\delta:M_n \rightarrow M_n \otimes M_n$ defined by $\delta(A_{ij}) := A_{ij} \otimes A_{ij}$ for some orthonormal basis $(A_{ij})_{i,j=1,...,n}$ of $M_n$ is necessarily a special commutative $\dagger$-Frobenius algebra in the category $\operatorname{fHilb}$ of finite-dimensional Hilbert spaces and complex linear maps, with co-unit $\epsilon: M_n \rightarrow \mathbb{C}$ defined by setting $\epsilon(A_{ij}) := 1$ (together with their respective adjoints $\delta^\dagger$ and $\epsilon^\dagger$). If $\delta$ is a CP map, then so is $\epsilon$, and $(\delta,\epsilon,\delta^\dagger,\epsilon^\dagger)$ is also a special commutative $\dagger$-Frobenius algebra in the category $\operatorname{CPM}(\operatorname{fHilb})$ of finite-dimensional Hilbert spaces and completely positive maps. 
The point is that every isometric comonoid in $\operatorname{CPM}(\operatorname{fHilb})$, such as $(\delta,\epsilon)$ would be when $\delta$ is CP, must necessarily involve pure maps $\delta,\epsilon$. This is because of the purity principle: whenever for pure CP maps $\Psi_i$ and $F$ we have $\sum_i \Psi_i = F$, necessarily $\Psi_i = q_i F$ for all $i$ and some coefficients $q_i \in \mathbb{R}^+$ (not necessarily all non-zero). The broad strokes of the proof are as follows.

The isometry condition $\delta^\dagger \circ \delta = id$ implies that $\delta$ must be $\mathbb{R}^+$-linear combination (not necessarily convex) $\delta = \sum_i q_i V_i$ of isometries $V_i$ with pairwise orthogonal ranges (i.e. $V_j^\dagger \circ V_i$ is the identity $id$ for $i=j$ and the zero map for $i\neq j$).
The unit laws $(id \otimes \epsilon)\circ \delta = id = (\epsilon \otimes id)\circ \delta $ imply that $\epsilon$ must be pure.
By post-composing the associativity law with $\epsilon$ in all three possible ways, an associativity law for the isometries can be derived.
By post-composing the associativity law for isometries with $\epsilon$, all isometries $V_i$ are identified, and orthogonality of ranges proves that $\delta$ must in fact be pure.
Because $\delta$ and $\epsilon$ are pure, i.e. they arise from the comultiplication and counit of a special commutative $\dagger$-Frobenius algebra of $\operatorname{fHilb}$. Those are exactly the examples given by taking matrices $A_{ij} := |i\rangle\langle j|$ with a single entry one and the rest zero in some fixed orthonormal basis $\big(|i\rangle\big)_{i=1,...,n}$ of $\mathbb{C}^n$.