I’ve seen it stated that the sigma baryon, $\Sigma^0$, only decays to $\Lambda^0 \gamma$, and then $\Lambda^0$ decays to $p\pi^{-}$ or $n\pi^0$.
Correct; the first decay preserves strangeness, so it is electromagnetic, the second one violates strangeness, so it is weak.
But why does emitting a photon change the isospin of $\Sigma^0$ from 1 to 0, i.e. from $\Sigma^0$ to $\Lambda^0$? Also, I assume this is just because there are no good options, but why is the decay $\Sigma^0 \rightarrow \Lambda^0 \gamma$ 100% (I’ve seen it stated that it is 100%, but I want to assume it is really $\sim $ 100%)?
The (strong) isospin of $\Sigma^0$ is $I=1, I_3=0$, of $\Lambda$ is $I=0$, and the photon has both isoscalar and isovector pieces, since electromagnetism violates isospin. The BR is almost 100%, as the other channels, 2γ or $e^+e^-$ , are very suppressed, as they must be. There is no need for weak channels.
The weak decays $\Lambda^0 \rightarrow p\pi^{-}$ or $n\pi^0$ violate (strong) isospin by 1/2, the celebrated dominant $\Delta I=1/2$ piece of the weak hamiltonian. There is no paradox, if you imagined one.
At the quark level, the weak Hamiltonian breaks weak isospin, but there might be "accidental/incidental" preservation of $T_3$, which, however, as you must appreciate, cannot really be well-defined for hadrons which also contain right-handed quarks, weak isospin singlets. You might be tempted to pull limited technical stunts, but these should be avoided unless you are in full control.
