How many parallelepipeds can be constructed with vertices $(0,0,0), (0,0,1), (0,1,0), (1,0,0)?$ My friend showed this problem to me after her professor asked this question as a warmup. Most of my issues come with double- or triple-counting the parallelepipeds. The simplest example of this is the cube, you can take a square in the XY, YZ, or XZ planes and translate them $1$ unit, but they're all the same parallelopiped.
I grouped the parallelepipeds into those with square bases (I counted ten), $\sqrt 2$ x $1$ bases (I counted twelve, but I'm not sure about double/triple counting here), and what I call "slanted" bases with the base parallelogram spanned by $(0,0,1), (0,1,0), (1,0,0)$, and a fourth vertex. I've found at least two of this kind which are unique. I'm not sure how to multiply by axes in this case because again repeats are annoying.
Not too sure how to proceed from here. I have a feeling this is a graph theory question under the hood somehow. I thought the problem was really interesting though.
Here's most of my working out so far:
 A: Since angles don't matter, you might as well transform the space so that
each of the four points is equidistant from the other three.
That is, transform the four points to vertices of a regular tetrahedron.
This will help you count things symmetrically.
Now consider how many faces the tetrahedron shares with a parallelepiped constructed on its vertices.
You can share three faces, two faces, one face, or none.
If you share three faces, once you select the three faces there is only one way to complete the parallelepiped.
For two faces, once you select the two faces you have two choices how to add a fourth vertex to one face, but once you make that choice the fourth vertex for the other face is forced. (You actually have two ways to add a fourth vertex to the second face such that you get a perallelpiped in the end, but one of those ways results in three faces of the tetrahedron shared with the parallelepiped, so we consider it already counted.)
For one face, you have three ways to add a fourth vertex, but once you have done that there is only one way to construct the opposite face of the perallelpiped without sharing at least one more face between the two objects.
And it is also possible to construct a parallelpiped that uses all four vertices but shares no face with the tetrahedron. Consider how you can create a tetrahedron from four vertices of a cube without using three vertices from any single face of the cube.

Alternatively, I think you can do this as a graph-theory question
(just as you suspected).
Take four vertices corresponding to the four given points.
Connect a pair of vertices if and only if those vertices are endpoints of the same edge of the parallelpiped.
Then each "share three faces" case produces a graph with one central vertex and an edge from the central vertex to each other vertex.
Each "share two faces" case produces a connected graph that passes through the four vertices in sequence.
"Share one face" produces a graph with three vertices and two edges in one component and a single vertex in the other.
"Share no faces" produces a graph with no edges.
Notice that it is not possible for the graph to have a component with just two vertices. If you had a component with just two vertices, the four vertices of the parallelpiped connected to those two could not be vertices of the tetrahedron;
but that leaves only two vertices, which lie on an edge parallel to the first two connected vertices and therefore cannot be the other two vertices of the tetrahedron.
A: A parallelepiped with one vertex at the origin is determined by an unordered set of three non-coplanar vectors $u, v, w$.  It has eight corners:  $0, u, v, w, u+v, u+w, v+w$, and $u+v+w$.  We can enumerate all parallelepipeds whose vertices include $0, e_1 = (1, 0, 0), e_2 = (0, 1, 0),$ and $e_3 = (1, 0, 0)$ by looking at all possible ways to assign the three $e_i$'s to the seven nonzero corners and solving for $u, v$, and $w$ in each case.  I'll leave the "solving for $u, v, w$" step as an exercise; there's always exactly one solution unless our choices of $e_1, e_2,$ and $e_3$ are linearly dependent.
To make the casework a little more manageable, let's say $u, v,$ and $w$ (i.e. the corners adjacent to $0$) are type 1 corners, $u+v, u+w$, and $v+w$ are type 2, and the opposite corner $u+v+w$ is type 3.  We then split into cases based on the types of the corners $e_1, e_2$, and $e_3$.  In each case, relabeling $u, v,$ and $w$ doesn't change the parallelepiped (so we can name them in any order we choose), but relabeling $e_1, e_2,$ and $e_3$ corresponds to rotating or reflecting the parallelepiped in space (so we must count how many of these rotations and reflections are genuinely different).  I'll give one example of each possible configuration and then say how many different rotations and reflections it has.
Types 1, 1, 1 $\implies e_1 = u, e_2 = v, e_3 = w$.  1 parallelepiped (a cube), with rotational and reflective symmetry.
Types 1, 1, 2 $\implies e_1 = u, e_2 = v, e_3 = u+w$.  6 parallelepipeds including rotations and reflections.  Note that $e_1 = u, e_2 = v, e_3 = u+v$ doesn't work because these are linearly dependent.
Types 1, 2, 2 $\implies$

*

*Case 1:  $e_1 = u, e_2 = u+v, e_3 = u+w$.  3 parallelepipeds including rotations; they have reflective symmetry.

*Case 2:  $e_1 = u, e_2 = u+v, e_3 = v+w$.  6 parallelepipeds including rotations and reflections.

Types 2, 2, 2 $\implies e_1 = u+v, e_2 = u+w, e_3 = v+w$.  1 parallelepiped, with rotational and reflective symmetry.  This is the only one whose vertices don't have integer coordinates:  $u, v$, and $w$ are the three permutations of $(1/2, 1/2, -1/2)$.
Types 1, 1, 3 $\implies e_1 = u, e_2 = v, e_3 = u+v+w$.  3 parallelepipeds including rotations; they have reflective symmetry.
Types 1, 2, 3 $\implies e_1 = u, e_2 = u+v, e_3 = u+v+w$.  6 parallelepipeds including rotations and reflections.  Note that $e_1 = u, e_2 = v+w, e_3 = u+v+w$ doesn't work because these are linearly dependent.
Types 2, 2, 3 $\implies e_1 = u+v, e_2 = u+w, e_3 = u+v+w$.  3 parallelepipeds including rotations; they have reflective symmetry.
In total, that's 29 different parallelepipeds.
