Suppose we have a curve $C$ of genus two over an algebraically closed field of characteristic zero with a given Weierstrass point $p_0$, furthermore we assume this curve to have real multiplication by $\mathbb{Z}[(1+\sqrt{5})/2]$ i.e. there is an embedding of $\mathbb{Z}[(1+\sqrt{5})/2]$ into $\text{End}(\text{Jac}(C))$. The given Weierstrass point gives us a canonical choice for the polarization on $\text{Jac}(C)$, namely $\{q - p_0:q \in C\}$. Now suppose we have another such curve, I would like to know when we consider two such curves isomorphic. I guess it is clear that we want $C$ and $C'$ to be isomorphic and we want that an isomorphism sends the given Weierstrass point on the first curve to the given Weierstrass point on the second curve. We want this isomorphism to respect the real multiplication on the Jacobians. However, I don't know how to make this formal.
A way to induce the real multiplication on the Jacobian is by considering the Kummer surface $K(C)$ of the Jacobian. If $E$ is a twisted cubic on $K(C)$ intersecting six of the nodes of $K(C)$, then the preimage $H$ of $E$ on $\text{Jac}(C)$ has the property that
$ \lambda^{-1} \circ \phi_H \in \text{End}(\text{Jac}(C)) $
induces the embedding of $\mathbb{Z}[(1+\sqrt{5})/2]$ into $\text{End}(\text{Jac}(C))$. So it would be natural to ask that the induced map on Jacobians respects this divisor. It would be nice to be able to make the real multiplication more intrinsic to $C$, I've got a funny feeling it has something to do with orderings of the Weierstrass points, but I can't make it formal. Any help will be appreciated!