Do two Miller planes need to have similar arrangement of atoms?
In answer to you first question, no if hkl differ, yes if they do not. I try to explain below.

In a crystal the unit cell defines the repeating unit. Inside the unit cell the atoms are arranged as they are in the molecular structure with the molecule being at the same angle and position within each of the unit cells. Each atom scatters the x-ray radiation depending on how many electrons the atom has and on the position the atom has in the unit cell. The many repeating identical unit cell structures cause constructive interference and so spots are observed on the x-ray detector. Thus the structure is determined from the intensity of the spots on the detector not their position.

The hkl planes are a construction we devise to allow determination of the x-ray amplitudes because the constructive interference that causes scattering has to occur from atoms in the repeated positions as determined by the unit cell. (There are thousands to millions of unit cells in a crystal). Thus we define planes starting at  hkl = (0 0 0) and continue to larger values such as  (1 2 0 ) etc. etc. (100 50 1) etc. (Negative values have a line above the number $\bar 3$ for example.) 

Suppose we choose the plane (1 2 3), this is  parallel to (2 4 6) to (4 8 12) and so on and these planes are separated by a distance we will call $d_{hkl}$. We can reduce all these  planes to (1 2 3) and the reason for this is that the phase of the x-ray wave from plane (2 4 6) is $2\pi$ from that of (1 2 3) and that of (4 8 12) is $4\pi$ shifted. But as the phase repeats itself over $2\pi$ all these larger phases are simply the same as $2\pi$. A phase change of $2\pi$ is equivalent to a path difference of $\lambda$ or a whole wavelength, which might be easier to envisage as then waves add together exactly.

Now, if an atom is found bang on the plane (1 2 3) it will have a phase of $2\pi$ (or path-length change of $\lambda$) with respect to the origin so has the maximum scattering intensity possible for that plane. However, consider that the atom in not on the (1 2 3) plane, but at a position $x$ away (and so between two (1 2 3) planes) then the phase it has is now $\displaystyle \frac{x}{d_{hkl}}2\pi $ and so the scattered x-ray waves are out of phase by this amount and so the scattered amplitude is reduced. Now, of course, every the atom in the molecule will be at some distance from one of the parallel (1 2 3) planes and so will have its own particular scattering amplitude to add to the total for that plane. 

The same reasoning applies to each of the unique hkl planes and to each atom. It is the un-ravelling of these amplitudes that produces the x-ray diffraction structure and it is made more complicated because the x-ray detectors measure the intensity of the scattered light (which is  amplitude squared (as $I^*I$)) which means that the phase information is lost. This causes great difficulties in solving the x-ray structure.