For the case of n = 2, one seems to need 3 points. For with 2 points (the diagonal end points), there are infinite rectangles that can be constructed.

For the case of n = 3, any two adjacent (touching) rectangular faces of a cuboid seem to suffice. So, we have two adjacent/touching (n=2) rectangles. Let their common edge determine 2 points. Applying the result from above, we then need one extra point to uniquely determine each of the two touching rectangles. So, we seem to need 4 points.

Does this generalize to arbitrary n wherein we need n+1 points? Is there a proof of such a result?