Please be kind. I've been working on this for a long time and can't find an answer. Feel free to edit for clarity if you think the question can be better worded.
Background
It may help to see a previous question I asked about quantum channels in order to understand the basis for this question.
A quantum channel is a mapping between algebras of bounded linear operators on Hilbert spaces, $\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$, where $L(\mathcal{H}_{i})$ is the family of operators on $\mathcal{H}_{i}$. In general, we are interested in completely positive, trace-preserving (CPTP) maps. The operator spaces can be interpreted as $C^{*}$-algebras (with the involution being the standard Hilbert space adjoint, denoted by $\dagger$) and thus we can also view the channel as a mapping between $C^{*}$-algebras, $\Phi : \mathcal{A} \to \mathcal{B}$. Since quantum channels can carry classical information as well, we could write such a combination as $\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$ where $C(X)$ is the space of continuous functions on some set $X$ and is also a $C^{*}$-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between $C^{*}$-algebras. Note, however, that these are not necessarily the same $C^{*}$-algebras. Since the channels are represented by square matrices, the input and output $C^{*}$-algebras must have the same dimension, $d$ (i.e. physicists will often "cheat" and refer to the dimension of an $n\times n$ matrix as simply $n$). Thus we can consider them both subsets of some $d$-dimensional $C^{*}$-algebra, $\mathcal{C}$, i.e. $\mathcal{A} \subset \mathcal{C}$ and $\mathcal{B} \subset \mathcal{C}$. Thus a quantum channel is a mapping from $\mathcal{C}$ to itself.
A quantum channel may be written as a Kraus decomposition,
$T(\rho) = \sum_{i}A_{i}\rho A_{i}^{\dagger}$
where the $\left\{A_{i}\right\}$ are the Kraus operators (and square matrices) and where
$\sum_{i}A_{i}^{\dagger}A_{i}=\textbf{1}$
and $T(\textbf{1})=\textbf{1}$.
Suppose we have two quantum channels, $r$ and $t$
$\begin{eqnarray*}
r: \rho \to \sigma &
\qquad \textrm{where} \qquad &
\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\
t: \sigma \to \tau &
\qquad \textrm{where} \qquad &
\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}
\end{eqnarray*}$
where the $\left\{A_{i}\right\}$ and $\left\{B_{i}\right\}$ are the Kraus operators for the channels respectively. We form the composite $t \circ r: \rho \to \tau$ where
$\begin{align}
\tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\
& = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\
& = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag
\end{align}$
where $i \cdot j = k$. Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. $\sum_{k} C_{k}^{\dagger}C_{k}=\textbf{1}$.
Core argument
From above, we note that
$\tau=\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}$.
Suppose we only have two Kraus operators for each, i.e. $A_{1}, A_{2}, B_{1}, B_{2}$. Then
$\tau=B_{1}A_{1}\rho A_{1}^{\dagger}B_{1}^{\dagger} + B_{2}A_{1}\rho A_{1}^{\dagger}B_{2}^{\dagger} + B_{1}A_{2}\rho A_{2}^{\dagger}B_{1}^{\dagger} + B_{2}A_{2}\rho A_{2}^{\dagger}B_{2}^{\dagger}$.
($\tau$ of course has a matrix representation (in fact it is a square matrix representation). The following has nothing to do with the size of the matrix representation of $\tau$ and only has to do with the terms in the above summation.)
Using the subscripts as a guide, I can make a matrix
$\begin{equation*}
\left(
\begin{array}{c c}
B_{1}A_{1}\rho A_{1}^{\dagger}B_{1}^{\dagger} & B_{1}A_{2}\rho A_{2}^{\dagger}B_{1}^{\dagger} \\[8pt]
B_{2}A_{1}\rho A_{1}^{\dagger}B_{2}^{\dagger} & B_{2}A_{2}\rho A_{2}^{\dagger}B_{2}^{\dagger} \\[8pt]
\end{array}
\right).
\end{equation*}$
This just happens to be the same dimension as the matrix representation of $\sigma \otimes \rho$. If I then do repeated composition I get,
$\begin{equation*}
\left(
\begin{array}{c c c}
B_{1}A_{1}\rho A_{1}^{\dagger}B_{1}^{\dagger} & B_{1}A_{2}\rho A_{2}^{\dagger}B_{1}^{\dagger} & \cdots \\[8pt]
B_{2}A_{1}\rho A_{1}^{\dagger}B_{2}^{\dagger} & B_{2}A_{2}\rho A_{2}^{\dagger}B_{2}^{\dagger} & \cdots \\[8pt]
\vdots & \vdots & \ddots
\end{array}
\right).
\end{equation*}$
The next step is simply to clarify the purpose. Suppose now that I take the output of a quantum channel and feed it back in on itself. In this case, $\left\{A_{i}\right\}=\left\{B_{i}\right\}$. Thus if we repeatedly apply the same channel $n$ times,
$\begin{equation}
T(\rho) \circ T(\rho) \circ \cdots \circ T(\rho) = \sum_{i^{n}}(A_{i})^{n}\rho (A_{i}^{\dagger})^{n}
\end{equation}$
we can take the terms of this expansion, form a matrix out of it, and that matrix (which may or may not have any physical significance) turns out to have the same dimension as the matrix representation of,
$T(\rho) \otimes T(\rho) \otimes \cdots \otimes T(\rho) = \bigotimes^{n}_{i=1} T(\rho)$.
Physically, the last equation is like applying $n$ copies of a channel simultaneously. In other words, there may be some kind of strange physical link between applying $n$ copies of a channel simultaneously and applying them in succession.
Questions in brief summary: Basically (and you can read the specific questions below) I need to know if a) the math for the core argument is right and b) what the immediate algebraic implications of it are (if there are any).
Question 1: The obvious question is, is this right, i.e. does the composition of quantum channels really have a representation that is morphic to a tensor product of quantum channels of a certain dimension (or is this obvious to a pure mathematician)? Is this generally true at the level of category theory?
Question 2: If it is right, the question that naturally follows is, what are the immediate algebraic implications (if any)?
Sub-question 2a: What kind of morphism is this on the level of monoids and/or categories, i.e. is it an isomorphism, epimorphism, etc.?
Best Answer
I think the answer is "sort of". A mathematical way to think about the "Kraus" operator is as follows. Set C = B(H) (with H finite-dimensional if you wish). Then, assuming the sum in the Kraus operator is finite (again, it could be infinite if you wish) then we can define a map $A:H \rightarrow H^n$ by a "column" operator $$A(x) = ( A_i^\dagger(x) )_{i=1}^n\qquad (x\in H).$$ Then your Kraus operator is $$ T(\rho) = \sum_i A_i \rho A_i^\dagger = A^\dagger (\rho\otimes 1) A. $$ Here $\rho\otimes 1$ is the operator on $H^n$ given by applying $\rho$ to each coordinate: the notation is explained by observing that $H^n$ is the Hilbert space tensor product $H\otimes \ell^2_n$.
So, if you have another operator $S$ given by $(B_j)_{j=1}^m$ we can form $B(x) = (B_j^\dagger(x))$ and then $$ S(T(\rho)) = B^\dagger(T(\rho)\otimes 1)B = B^\dagger (A^\dagger\otimes 1)(\rho\otimes 1\otimes 1)(A\otimes 1)B = C^\dagger (\rho\otimes 1) C,$$ where $C = (A\otimes 1)B : H\rightarrow H^{nm} = H \otimes \ell^2_n \otimes \ell^2_m$.
So, you see why tensors appear. But this is a bit different to what you had...