If we assume that the lattices have the same structure, the lattice enthalpy is dominated by the charge of the ions and the distance between them. Since we are dealing here with cations from the same group, the charges are the same, so the remaining variable is the distance between them.
With moderately sized or small anions, a smaller cation can get closer, resulting in a shorter overall distance and a larger magnitude lattice enthalpy. In the case of very large anions, however, the anions are already packed as closely together as possible, so if a smaller cation moves closer to one sulfate, it moves farther from another. The equilbrium position of the cation is thus determined solely by the distance between the sulfates and does not vary with cation size.
For the second part of your question, I'm guessing that your statement that "larger cations stablise larger anions and small cations stabilise small anions" might be made with respect to solubility?
Solubility is a function of both the lattice enthalpy and the solvation enthalpy, which are in competition with each other. Generally speaking, small ions have higher solvation enthalpy in water. However, they also can have greater magnitude lattice enthalpy if they are able to pack more closely together.
Empirically, it has been observed that if the cation and anion are both small, the lattice enthalpy "wins" over the solvation enthalpy, and the salts tend to have low solubility in water.
If both ions are large, both the lattice enthalpy and the solvation enthalpy have smaller magnitude, and the empirical observation is again that the lattice enthalpy "wins".
When the sizes are mismatched, however, we find that the solvation enthalpy "wins" and the salts are generally soluble. We can see why in the extreme example from the first part of your question, where a smaller cation does not increase the magnitude of the lattice enthalpy at all, but will definitely increase the magnitude of the solvation enthalpy relative to a similar salt with a larger cation.
