The $\Lambda^0$ and $\Sigma^0$ have respectively I=0 and I=1. The source of your confusion may come from the fact that the electromagnetism doesn't respect the isospin symmetry. In E.M. processes, only $I_3$, the third component, is conserved. The total isospin $I$ is not. A trivial example is the neutral pion decay $\pi^0 \to \gamma \gamma$. The pion is a member of an isospin triplet $(\pi^+,\pi^0,\pi^-)$ (thus $I=1$ and $I_3=0$) while photons cannot carry any isospin numbers (since not made of $u$ or $d$ quarks).
Edit (to answer a comment): how do we know that $\Sigma^0$ has $I=1$?
There are several experimental facts. The $\Sigma^+,~\Sigma^0,~\Sigma^-$ have almost the same mass (about 1190 MeV). So historically, they were considered in the same multiplet of isospin, reasonably a triplet as for $\pi^+,~\pi^0,~\pi^-$. In addition, the reaction $K^- + p \to \Sigma^0$ is seen (strong interaction). Knowing that $p$ has $I=1/2,~I_3=+1/2$ and $K^-$ has $I=1/2,~I_3=-1/2$, the initial state $K^-+p$ can have $I=0$ or $I=1$ (sum of 2 Isospin 1/2). If you admit that $\Sigma^0$ is a member of a multiplet (previous argument), you're forced to conclude that $I=1$. I'm sure that many other reactions can be found in the literature justifying $I=1$.
In the modern quark languages, $\Sigma^+ \equiv uud,~\Sigma^0\equiv uds,~\Sigma^-\equiv dds$. With the group theory, assuming the (approximate) SU(3) flavor symmetry, $I=1$ comes "naturally".
Isospin is a made-up concept to explain the interactions between the neutron and proton via pions. Effectively there is a certain symmetry that is explained by the SU(2) group nicely. You can use it to predict branching ratios and hence rates of decays.
We assign isospin of +1/2 to u quark and -1/2 to d quark. You can look into Young's tableaux and derive 2 ⊗ 2¯ = 1 ⊕ 3. That means that if you combine a quark and an antiquark you get 1 symmetric and 3 antisymmetric states (much like the singlet and triplet that you get in the S and P shells of Helium. You can have both electrons pointing up,down or somewhere in the x,y plane in the same direction, which would always give S=1, or they can be pointing in different directions whichever axis you pick S=0)
Now the triplet has I=1 and Iz=-1,0,1 and can easily be identified as the 3 pions. The singlet is a bit more tricky as you need the s-quark to make a nonet of mesons to identify a singlet.
Best Answer
The decays involving photons are, of course, electromagnetic. You may forget about weak intermediaries.
The e.m. current coupling to the photon in the lagrangian in terms of the lightest quarks is $$eA_μ(2/3\bar{u}γ^μu−1/3\bar{d}γ^μ d)=eA_\mu \bar{q}(\tfrac{1}{6} \mathbb {1}+I_3)\gamma^\mu q,$$ since diag(2/3,-1/3)=diag(1/6,1/6) + diag( 1/2,-1/2). Each quark couples to the photon through its (different) charge. So the photon couples to both isoscalars (I=0) and isovectors (I=1), and electromagnetism violates strong isospin. However, if you imagine this coupling term to preserve strong isospin, then you must represent the photon as a linear combination of (strong) isoscalar and isovector.
As a consequence, if you had to preserve strong isospin, as a way to track its violation, both vertices $\eta \gamma \gamma$ and $\pi^0 \gamma \gamma$ are allowed (they both have isospin zero) in this accounting, and, indeed, both particles decay to two γs as specified by the vertices.
Considerations of weak isospin, spontaneously broken anyway, are irrelevant in these decays.