If such a pyramid exists, could someone provide the coordinates of its vertices?
If by pyramid, you mean the shape with a square base and four triangular sides, and you want the right angles all where the triangles meet, there is a solution, but it is degenerate.
Sample Vertices: Set A: (0,0,0), (1,0,0), (0,1,0), (-1,0,0), (0,-1,0) Set B: (0,0,0), (1,1,0), (1,-1,0), (-1,-1,0), (-1,1,0)
By degenerate, I mean that all of the vertices are in the same plane (z=0, in my example), and the shape has zero volume. It's flat.
It is obvious that this must be the case. You have four right angles meeting at a point. That's 360 degrees, exactly what you need for them to be in the same plane.
If the first point in my example A was (0,0,1), then all of the edges would be the same length (the square root of 2), so the triangles would all be equilateral--and all of their angles would be 60 degrees. The more you raise that point, the smaller the angles there get.