Actually, I think the channel is not extremal because I suspect you are misquoting the Landau-Streater result. So I will state it here.

To be precise, for anyone unfamiliar with the field, a *quantum channel* is a trace-preserving, completely-positive linear map on density matrices (positive semidefinite matrices with unit trace), of potentially different sizes. A basic theorem in quantum information says that every quantum channel from $m\times m$-dimensional to $n\times n$-dimensional density matrices can be written in *Kraus form*:
$$ \rho \mapsto \sum_{i=1}^N A_i \rho A_i^\dagger, \text{ for linear operators } A_k \colon \mathbb{C}^m \to \mathbb{C}^n \text{ satisfying } \sum_k A_k^\dagger A_k = I_m. $$

It is easy to show that the set of quantum channels between systems of fixed dimension is convex. It also easy to show that the set of channels that map $\frac{1}{m} I_m$ to a fixed density matrix $\sigma$ is convex. Now the theorem of Landau-Streater says that if $m = n$, a channel with Kraus form as above is extremal in this latter set if and only if the $N^2$ linear operators $A_i^\dagger A_j \oplus A_j A^\dagger_i$ (of size $2m \times 2m$) are linearly independent. It seems you have instead been working with $m\times m$ matrices. But I think that even if you were to continue and apply the theorem correctly, you would only prove or disprove extremality in the convex subset of unital channels, i.e. those for which $\frac 1m I$ is a fixed point. So potentially you could strengthen Ben-Or's conclusion by showing non-extremality in this subset, or otherwise you might conclude extremality there, which would tell you nothing about extremality in the entire set of channels.