$Sigma^0$ and $n$ decay

You may be basically asked to understand the PDG. You can't change strangeness, electromagnetically.

So you may only decay by emitting a photon and rearranging your quarks in the case where your baryon charge stays the same (!), $Sigma ^0 to Lambda gamma$, which makes the neutral Σ nine orders of magnitude shorter-lived than its isopartners. No other options are available, energetically, and this is the only spot in the octet where this (same-charge members) happens. (And, as you appreciated, you must conserve baryon number, so that option is closed.)

Now, in a different world, the neutron would prefer to decay (by rearranging quarks) to a proton and a $pi^-$ if it could, energetically: but check that the n-p mass difference is too small. So its only option is the usual weak decay to an electron and an antineutrino and a proton (semileptonic: some of the decay products are leptons) and it takes it forever to do that, about a quarter of an hour...

The $Sigma^-$, however, has that option, since it is so much heavier than the neutron, to decay weakly to $n~pi^-$, which it does most of the time. But your mode $n~e^- bar nu$ is also OK, except subdominant (BR ~ $10^{-3}$), and is, in fact, measured, Bourquin 1983, providing valuable WI information.