For the Ising model with only nearst neighbor interaction on square lattice, if we do the RG by integrating out half degree of freedom, then we would get a new Ising model with many kind of interactions, so Ising model with only nearst neighbor interaction cannot be a fixed point of RG.
In general, the fixed point should include infinite many kind of interactions and we cannot find it exactly.
But for now assume we find it, i.e., we have an Ising model with infinite many kind of interactions and it is a fixed point of RG, and we consider the two point spin-spin correlation,$\langle s(0)s(r)\rangle$. Before the RG, the distance for two spin is $r$, after the RG, the distance becomes $\frac{r}{2}$,but the Hamiltonian remains the same except the number of spins become half .So I think the $\langle s(0)s(r)\rangle=\langle s(0)s(\frac{r}{2})\rangle$. But obviously it is wrong since the spin-spin correlation function should decay as power law. What is wrong with my argument?