seventeen tiles on a torus The torus $\mathbb{R}^2 \mod((4,1),(1,-4))$ has area 17.  Can it be covered by seventeen labelled tiles in two different ways so that any pair of tiles is neighbours of each other (at an edge or a vertex) in one of the two patterns?
 A: With the generating translations as $(4,1),(-1,4)$ we see we have a square lattice tipped a bit from the axes. Note we can also move by $(-4,-1),(1,-4).$ We need a complete and nonredundant list of the translates of the (axes aligned) unit squares under the translation group. Label initially each unit square by the coordinates of its lower left corner. Then a nonredundant list is
$$(-1,3),\ (0,3),\ (1,3),\ (2,3),\  (3,3),\\
(0,2),\ (1,2),\ (2,2),\ (3,2),\\
(0,1),\ (1,1),\ (2,1),\ (3,1),\\
(0,0),\ (1,0),\ (2,0),\ (3,0).$$
Note that geometrically these unit squares make up a region consisting of a $4\times 4$ square with an extra unit square stuck on to its upper left, flush with the top of the big square. These shapes then tile the plane, as a drawing shows. The generators of the transformations of the group serve to move these shapes around to achieve the tiling.
Now for ease of reference relabel these as the integer sequence $1,2,\cdots,17$ going across the rows one by one, so that e.g. $(-1,3)$ gets the new name $1$, and to take another random case $(1,2)$ becomes $7.$
In terms of these numeric labels, regarded mod 17, the point $x$ and its neighbors are the nine points
$$x+12,x+13,x+14,\\ x+16,x,x+1,\\x+3,x+4,x+5.$$
These formulas come from using the translations of the group to see which identified unit squares end up bordering a fixed one of them, and the above arrangement is in the shape of the $9 \times 9$ square containing the particular unit square $x$ being looked at.
The list of addends in these formulas (for the neighbors of $x$) is the following subset of $\mathbb{Z}_{17}^*:$
$$A=\{1,3,4,5,12,13,14,16\}.$$
Note that when  $A$ is mapped by $f(x)=2x$ (mod $17$), its image is the complement of $A$ in $\mathbb{Z}_{17}^*$. It follows that if we take the map $f(x)=2x$ (mod $17$) and apply it to the labels in the "first tiling" $T_1$ above, we obtain a distinct (ordered) tiling $T_2$ and two numbered tiles touch (at corner or edge) in $T_1$ if and only if they do not touch in $T_2$.
A: I've done an exhaustive search for solutions.  I'll start with labels as follows:
 0
 1, 2, 3, 4
 5, 6, 7, 8
 9,10,11,12
13,14,15,16

(So this is immediately slightly different from coffeemath's solution, which I don't understand.)  If we constrain the solution to have 0 in the same position, there are 4 solutions, closely related to each other:
Solution S0:
 0,
 5, 7, 1, 3,
13,15, 9,11,
 6, 8, 2, 4,
14,16,10,12

Solution S1:
 0,
 3,11, 4,12,
 1, 9, 2,10,
 7,15, 8,16,
 5,13, 6,14

Solution S2:
 0,
14, 6,13, 5,
16, 8,15, 7,
10, 2, 9, 1,
12, 4,11, 3

Solution S3:
 0,
12,10,16,14,
 4, 2, 8, 6,
11, 9,15,13,
 3, 1, 7, 5

These solutions are in inverse pairs: S0-S1 and S2-S3.
All these solutions are isohedral, in the sense that we can specify a relationship between pairs of labels before and after.  For example:
S0: Let x be a label, and y be the label immediately lower than it in the original layout.  In S0, place y two squares LEFT OF x.
This is enough to define the solution.  For S1, S2 and S3, replace 'LEFT OF' with 'BELOW', 'ABOVE' and 'RIGHT OF' respectively.  We can regard S0-S1 and S2-S3 as reflections of each other in the x-y diagonal.
As I mentioned, there are no solutions except this collection of 4 solutions (up to translations of the tiling).
