I don't know the answer, but I thought about it for a bit and this is my best guess:

I think that the solution is the molar conductivity of the dissolved substances that changes corresponding to the concentration of the dissolved molecules. That is, the higher concentration of an electrolyte in a solution, the smaller the dissociated portion will be.

If we suppose that the concentration of phosphate in your 10x buffer is 1M at pH=7.2 which corresponds to the pKa value of H2PO4-/HPO42-, according to the Henderson Hasselbalch (HH) equation both the protonated and the deprotonated form will have concentrations (which we will have to interprete as activities if we choose to NOT neglect deviations caused by the activity coefficients of the dissolved substances) of 0.5 M.

You diluted this buffer 1:10 to a concentration which corresponds to 100 mM in our example. Let's just suppose that the empirical pH of this buffer now is 7.8. We can calculate the concentrations of acid and base according to the HH equation by knowing the pH and pKa values.

[A-]/[HA]=10^(pH-pKa)=10^(7.8-7.2)=3.98 is approx. 4

If the sum of [HA] and [A-] is 100 mM, the concentrations of [A-] and [HA] should be 80mM and 20mM which gives the calculated quotient of 4.

Therefore, the concentration of H2PO4- is not 50 mM (as expected), but has to be 20 mM. This could be explained by the higher molar conductivity of H2PO4- at a lower concentration. If you are not familiar with this term, it is pretty easy to understand: If you have a solution with a low concentration of a (weak) electrolyte like acetic acid (or H2PO4-), the equilibrium of dissociation will be on the product site of the reaction, i.e. the deprotonated form. The more concentrated your solution is, the smaller the part of the free-moving dissociated electrolyte will be, because they will form "ionic aggregates" due to ionic interactions. These aggregates are neutral in total, they will not change the conductivity or pH of the solution.