How can I prepare two phosphate buffer of different concentrations but give same pH? How is it that buffers of different compositions can have the same pH? For example, it is possible to prepare 0.01 M phosphate buffer of pH 7.0 and 0.1 M phosphate buffer of pH 7.0? How?

I used Henderson-Hasselbalch equation, but still failed. No way is working. Is it something related to volume? Can it be shown mathematically? Can anyone show it mathematically that it is possible?
For a phosphate buffer with $\mathrm{pH} = 7,$ the two dominant species are $\ce{H2PO4-}$ and $\ce{HPO4^2-}$. The relevant $\mathrm{p}K_\mathrm{a}$ is $7.2$ (this is $\mathrm{p}K_\mathrm{a2}$). From the Henderson-Hasselbalch equation, using $\mathrm{pH} = 7,$ the ratio of $\ce{HPO4^2-}$ to $\ce{H2PO4-}$ is about $0.631.$ You can see this in the alpha diagram below:



In this figure, the green trace is the $\ce{H2PO4-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14.$ The blue trace is the $\ce{HPO4^2-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14$. So, at $\mathrm{pH} = 7,$ the blue curve is below the green curve, and the ratio of alpha fractions is $10^{-0.2}$, i.e., about $0.631.$ This is directly from the Henderson-Hasselbalch equation. Since all the phosphate species are in the same buffer volume, this is also the ratio of the respective concentrations.

The difference between the $\pu{0.01 M}$ phosphate buffer and the $\pu{0.1 M}$ phosphate buffer is that the $\pu{0.1 M}$ buffer is ten times higher in both $\ce{HPO4^2-}$ concentration and $\ce{H2PO4-}$ concentration. So the ratio is the same as in the $\pu{0.01 M}$ buffer and the $\mathrm{pH}$ is the same for the two buffers. Of course, the more dilute buffer has lower buffer capacity, but that is another issue.