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.

The weak decays $\Lambda^0 \rightarrow p\pi^{-}$ or $n\pi^0$ violate (strong) isospin by 1/2, the famous 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.

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.

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 $\Lambda$ $I=0$, 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.

The weak decays $\Sigma^0 \rightarrow p\pi^{-}$ or $n\pi^0$ violate (strong) isospin by 1/2, the famous dominant $\Delta I=1/2$ piece of the weak hamiltonian.

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.

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.

The weak decays $\Lambda^0 \rightarrow p\pi^{-}$ or $n\pi^0$ violate (strong) isospin by 1/2, the famous 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.