@Joel My apologies, I was indeed a bit sloppy. I've rephrased it in terms of forcing, which I find a more natural way of looking at it. If I'm not mistaken, Theorem 1 from your blog post implies that for almost equivalent complete BAs (under my rephrased definition), every condition has a cone equivalent to a cone from the other forcing.

You must be reading my mind by responding to my question within 6 seconds...

Thank you! Does this work in the other direction, i.e., if we assume $\kappa$ is $\lambda$-supercompact, then does my desired property hold?