Thoughts on Extending the old Fair Partitions Question

Recording some leads from the original fair partition (spicy chicken} question which asks for partitions of ocnvex bodies into convex pieces of same area and perimeter:

1. Are the properties of the space of partitions into n pieces with 3 quantities equal among pieces (rather than 2) well understood (the equalized quantities could be say, area, diameter, perimeter)?

Response from Prof. Roman Karasev: one may first invoke the configuration space of regular convex partitions into $n$ parts, which has dimension $3n-4$. But the number of equations to solve is $3n-3$. Hence there is a problem in dimension counting already. Of course, not all convex partitions are regular and there are in fact more degrees of freedom. But from the point of view of equivariant topology the space of all convex partitions into $n$ parts equivariantly retracts onto the configuration space of dimension $2n$ and then to an $(n-1)$-dimensional poyhedron. This roughly means that equalizing 2 values (one of them is area) is the best possible result attainable with equivariant methods.
------------
2. What are the possible conditions under which the pieces resulting from a partition guaranteed to be congruent? For example, can one have claims like (say):
"if, for 2 values of n (n1 and n2) that are relative prime, a given convex region can be cut into n1 pieces all with 3 properties equal and also into n2 pieces with the same 3 properties equal among pieces (but not into numbers of pieces which are not relatively prime), then, the pieces resulting from both partitions are necessarily congruent"?

The best educated guess for the above quesiton that I heard is that there is no way to guarantee congruence from any finite set of inequalities - iow, there is sufficient perturbative freedom left so that one can cook up a convex region C that allows such partitions into relatively prime numbers of pieces not all congruent but with all the required quantities equal across pieces.
------------
3. Even if only 2 quantities need to be equal among the n convex pieces, if one of them is NOT area, then, things are different. Here is something I just learned from Prof. Karasev:


The current techniques allow to equalize two quantities among pieces when at least one of them behaves like area, that is tends to zero when a part of the partition tends to the empty set (note that this is not the case for perimeter). Moreover, the restriction that the number of parts $n$ is a prime power is also hard to avoid (unless one of the quantities is additive, like the area).

So, there is a chance that say "to equalize perimeter and diameter among 6 (smallest nonprime power) pieces may not be possible"!