
King Knight's Gambit 
There appear to be two knotty issues in attempting to find a general solution to the homometric partition puzzle. First is the obvious problem posed by ntuples. It is well known that zrelated sets don't only come in pairs; there are zrelated triples, quadruples, etc. This appears to be the most difficult problem. I'll put it on hold for now.
Second is a more subtle conjunctdisjunct problem. If two zrelated sets
S and
T can be rotated or reflected (transposed or inverted) so they are in positions disjunct from one another, then we can always find a partition solution by the basic partition theorem given in the previous post. However, there are many cases for which two zrelated sets cannot be positioned such that they are disjunct.
The easy case is when
S and
T have cardinalities larger than half the size of their
Kspace, but they are zrelated because their smaller complements are zrelated. A corollary to the basic partition theorem takes care of this type of case:
An example can be taken from
K12. S={0,1,4,5,7} and
T={3,6,9,10,11} are zrelated.
Q={2,8} and V(
S,Q)=V(
T,Q)=[0,2,2,2,2,2,0]. Their zrelated complements are
S'={2,3,6,8,9,10,11} &
T'={0,1,2,4,5,7,8}, and
Q'={0,1,3,4,5,6,7,9,10,11}. V(
S',Q')=V(
T',Q')=[5,12,12,12,12,12,5]. (Remember the first element of the spanning vector is the number of 0's (harmonic/melodic unisons, rhythmic simultaneities).)
When we get to
K13 we run into a situation that at first appears to defy any attempt to apply any variation of the approach taken by the HPT. There are just two pairs of zrelated hexads in
K13. {0,1,2,4,5,8} & {0,3,4,5,6,9} share the ICV [6323322] and {0,1,3,5,7,8} & {0,1,2,4,7,9} share ICV [6232233]. There are several notable things here.
K13 is the first
Kspace with an odd number of nodes in which zrelated sets appear. And 13 is not only odd, it's prime. Also, note the peculiar relationship between the vectors: their nonzero entries
appear to be related by exchanging the ic cardinalities 2 and 3. Even though you may know why these two pairs are related this way, please stick with me through the following exposition which will lead off on a path that, at least within music theory (to my knowledge), is untrodden.i
Recall that in K12 there is a sort of subcanonical atonal operation, usually denoted M5, which is simply multiplication by 5 mod 12. One of the interesting properties of this operation is that by modding back to 12 it exchanges interval classes 1 and 5, so if the ICV of some random set X is [aBcdeFg], the ICV of 5X will be [aFcdeBg], no matter how the operation may warp the shape of the set. And if B=F, X and 5X will also be Tn, TnI, or zrelated. In K12 this operation only "works" (i.e., yields a full chromatic cycle) for multiplication by 1, 5, 7, and 11 because these are the only integers available that are coprime with the modulus 12.
But since 13 is a prime number, in K13 all available integers 1 through 13 are coprime with 13. For example, if X={0,1,2,4,5,8}, then 2X(mod13)={0,2,4,8,10,3}, so that V(X)=[6323322] and V(2X)=[6232233] (Figure 1).

Figure 1 
So there isn't really an exchange of components between the vectors of these two pairs of sets, but a shuffle permutation that, in this example, gives a firstblush impression of such an exchange.[1] All of this is familiar territory to seasoned music theorists, even those who have never considered this particular case in K13 before.
Now comes the breakout question that might help us generalize the basic partition theorem to any pair of homometric sets, whether or not they are disjunct, and whether or not they appear in an even or odd Kspace. Here's the question:
What if we don't mod out after multiplication?
In the example at hand, such a shift in procedure, rather than keeping us in K13, maps us into the subset consisting of only the evennumbered elements of K2X13 (i.e., K26). Furthermore, this shift retains the original ICV on the evennumbered components of the ICV of the transformed set(s) (Figure 2).

Figure 2 
To reiterate & make the next move more clear, switch to a pictorial description. Look at the inscribed hexagons
X={0,1,2,4,5,8} (the red hexagon) and
Y={0,3,4,5,6,9} (blue hexagon) situated in
K13 (Figure 3).

Figure 3 
To apply the HPT we would like to rotate or reflect one of these hexagons such that X and Y are disjunct, leaving a singleton set for Q. Unfortunately, there is no way to do this – unless we think outside the circle, so to speak. Multiplying by 2 (without modding back into K13) yields the situation shown in Figure 4.

Figure 4 
This gives us X'=2X={0,2,4,8,10,16} and Y'=2Y={0,6,8,10,12,18}.
The shapes of the hexagons in both cases are identical (invariant under a linear homothetic transformation). What has changed is that now they are situated only on the even numbered nodes, leaving us with an additional 13 nodes (the odd ones) to play with (Figure 5).

Figure 5 
Rotating or reflecting either hexagon to oddnumbered nodes will guarantee that the two are disjunct. So let's move the blue hexagon by one click clockwise (Figure 6)

Figure 6 
and ...
(1) we now have two hexagons with the same shape[2] as those we started with in K13;
(2) the K13 homometric relation is carried over into K26;
(3) the initial intervalclass vector shared by X and Y in K13 is retained in the even components of the K26 vector for X' and Y' as well as any transposition of Y' (Figure 2);
and, finally, what is gained by this transformation,
(4) X' & Y' are disjunct and the basic partition theorem HPT can be applied. And ... voilà!
For the situation shown in Figure 6, X'={0,2,4,8,10,16}, Y"=T1(Y')={1,7,9,11,13,19}, Q = compl(X'UY") = {3,5,6,12,14,15,17,18,20,21,22,23,24,25}, and by the BPT:
V(X',Q') = V(Y",Q') = [06658866658785].
Obviously, this can also work in reverse, given the right initial conditions.
If(
!) two homometrically related figures can be rotated so that they are both placed on the even nodes of a
K2nspace, whether that rotation makes them disjunct or conjunct, after the entire space is
divided by two, in the resulting
Knspace the original figures will appear retaining their same shape and homometric relationship, as well as the same vector (with the zeros removed).
☛ Note to self & others: Lack of formality is still making this whole edifice unstable. It's still in a sense a CIP, a conjectureinprocess.
_________________
* If the metaphorical connection of the title to the post content is unclear after reading, the answer will be provided in the next post.
[1] In passing, it's no secret that moddedout multiplication of pitch classes and interval classes can be alternatively expressed as permutations. Multiplication of ic's by 6 (mod 13) is summarized by the permutation (0)(615243), a spiral permutation in the same category as the sestina poetry form and the Klondike card shuffle. I'll be returning later to a few unusual but significant 2021st century examples of permutations as compositional tools.
[2]
A note from Charles Ives re escaping Diatony's gravitational pull: It may be helpful at first to understand "shape" as "chord." Example: If you have an augmented triad {0,4,8} in 12tET it can be described by its intervals expressed in semitones as (4,4,4). If you multiply by 2 mod 12, you get the same augmented triad {0,8,4}. Now, if you multiply by 2
without modding back into 12tET, you effectively still end up with a triad that sounds exactly the same as the initial {0,4,8} triad, but now "spelled" {0,8,16} in 24tET and described as an interval string using quarter tones, (8,8,8). If multiplyby2 is all you do, you gain nothing  the two triads sound the same whether you claim to be in 12tET or 24tET 
until a second augmented triad comes along that occupies 24tET's
oddnumbered nodes, say it's spelled {9,17,1}. At that point you have gained a trichord relationship with the voiceleading (spanning) vector [0300000303000]  and a hexachord ({0,1,8,9,16,17} w/string (1,7,1,7,1,7))  not possible in 12tET, even though the generating "
shapes" (augmented triads),
considered individually, are indistinguishable. This is just the general idea. I used much more complex relationships in an experiment a while back.