There are four (or five) two dimensional Bravais lattices which I refer to as oblique, rectangular, hexagonal and square.
I'll discuss the 17 symmetry wallpaper groups below.
Ignoring translation, if I ask someone to lay one hexagonal lattice on top of another with one of their axes parallel, there's only one distinct rotational way to do it. One might think there are six (since the lattice is p6 or six-fold rotationally symmetric) but they are indistinguishable.
If I ask for an offset of 10 degrees then there are two ways (+/- 10°).
If I ask for an offset of 60/2 = 30 degrees, the number of ways drops back to one.
The expression or algorithm I need would fill out the following table:
Oblique Rectangular Hexagonal Square
2 2 6 3
Oblique 2 1/1
Rectangular 2
Hexagonal 6 3/6 (30) 1/2 (30)
Square 4 1/2 (45)
This is read as follows: For hexagonal on hexagonal every 30 degrees there is 1 way, and for angles that are non-zero modulus 30 degrees there are 2 ways. For square on hexagonal the number is three and six, also every 30 degrees as shown here
Square and hexagonal lattices shown separately:
Square and hexagonal lattices shown at 0, 30 and 60 degrees:
Square and hexagonal lattices shown at +/-10 degrees (only two of the possible six are shown):
In surface science if we had islands of one lattice on top of another that different only in this way we'd call them domains.
Question: For pairs of 2D lattices each with either a 2, 3, 4 or 6-fold rotational symmetry (ten possible pairs in total), how can I calculate the number of possible ways I can uniquely configure them with a given angle who's absolute value is $\theta$.
This is of course likely to be a solved problem, so if one wants to point me to a description of the solution that is not written at a high level rather than doing so here, that would also be great!
Wallpaper
The problem is more complex if the specific symmetry group of each lattice is defined and if that can be included in the answer at the same time it will be even more helpful. I'm not sure how to formulate that question exactly so I've ask the question who's answer will at least get me started.