This is not a complete answer, but it might help.

The map $T: A_i \mapsto A_i \otimes A_i$ sends, by its very definition, the orthonormal family $(A_i)$ to an orthonormal family. It is therefore an isometry for the Hilbert-Schmidt norms.

But there are not that many completely positive maps $M_n \to M_m$ which are also isometries for the Hilbert-Schmidt-norms. Namely such a map is of the form $T(x)= D \pi(x)$, for some (not necessarily unital) $*$-homomorphism $\pi$ and some positive operator $D=T(1)\in M_m$ commuting with the range of $\pi$. This is an if-and-only-if condition provided that $\|D\|_{HS}=\sqrt n$. This statement is probably known. If you want I can expand the proof I have in mind.

This implies that such a map satisfies $Tr\circ T=c Tr$ for some positive $c=Tr(D)/n$, and more generally that for any $p>0$, $\|Tx\|_p = c_p \|x\|_p$ for $c_p=\|D\|_p/n^{1/p}$.

Coming back to your problem, I do not see how to conclude, you can already find a couple of necessary conditions on the $A_i$'s for the map $T(A_i)=A_i \otimes A_i$ to be completely positive.

Matthew asked for a proof of

A linear map $T:M_n\to M_m$ is completely positive and isometric for the Hilbert-Schmidt norm if and only if $T$ is of the form $T(x)= D \pi(x)$, for some (not necessarily unital) $*$-homomorphism $\pi$ and some positive operator $D=T(1)\in M_m$ commuting with the range of $\pi$, and such that $\|D\|_{HS}=\sqrt n$.

I only prove the "only if" direction. Assume that $T$ is cp and isometric for the Hilbert-Schmidt norm. Using the fact that $T$ is cp, by Stinespring's theorem, there is a (finite dimensional) Hilbert space $H$ and a linear map $V:\mathbb C^m \to H\otimes \mathbb C^n$ such that $T$ can be decomposed as $T(x)=V^* 1_H \otimes x V$. I claim that the assumption that $T$ is isometric implies that $VV^*$ is of the form $A \otimes 1_n$ for some positive $A \in B(H)$ (in particular $V V^*$ commutes with $1\otimes x$ for all $x \in M_n$). This will imply that $T(x) T(y) = T(1) T(xy) = T(xy) T(1)$ for all $x,y \in M_n$, and hence putting $\pi(x) = T(x) T(1)^{-1}$ (with the convention $0/0=0$) we get the proposition.

The claim is not complicated to check. By the trace property, $\langle Tx,Ty \rangle = Tr(VV^* (1\otimes x) V V^* (1\otimes y^* ))$. Writing $VV^* = \sum B_{i,j} \otimes e_{i,j}$, taking $x=e_{i,j}$, $y=e_{s,t}$, and using that $T$ preserves the scalar product, one gets $\langle e_{i,j},e_{s,t}\rangle= Tr(B_{s,i}B_{j,t})$. But $VV^*$ being self-adjoint, this becomes $\delta_{i,s}\delta_{j,t}= \langle B_{s,i},B_{t,j}\rangle$. This implies that $B_{s,i}=0$ if $s\neq i$ and that the matrices $B_{i,i}$ are all of Hilbert-Schmidt norm $1$, and that $\langle B_{i,i},B_{j,j}\rangle=1$. Thus (equality in Cauchy-Schwartz inequality), the $B_{i,i}$'s are all equal, to some matrix $U$. This proves the claim.