Let $$ X=\{(i_1,\ldots,i_{n-1}) : i_j\in[1,n]\}. $$ Is there a "natural" way to decompose $X=\bigcup_kX_k$ such that for $x\in X_k$, no coordinate of $x$ is equal to $k$?
For example:
- [$n=2$] $X_1=\{(2)\}$ and $X_2=\{(1)\}$, no choices.
- [$n=3$] There are three equal-sized partitions (switching where $(1,1)$, $(2,2)$, and $(3,3)$ go), \begin{align*} X_1&=\{(2,2),(2,3),(3,2)\},\\ X_2&=\{(3,3),(1,3),(3,1)\},\\ X_3&=\{(1,1),(1,2),(2,1)\}, \end{align*} and other decompositions, e.g. greedy \begin{align*} X_1&=\{(2,2),(2,3),(3,2),(3,3)\},\\ X_2&=\{(1,1),(1,3),(3,1)\},\\ X_3&=\{(1,2),(2,1)\}. \end{align*}
I assume this is a combinatorial problem people have considered, but I can't think of a "natural" solution (note that such a decomposition always exists at the very least).
Motivated by expanding the product $$ (x^{(1)}_1+\ldots+x^{(1)}_n)\cdot\ldots\cdot(x^{(n-1)}_1+\ldots+x^{(n-1)}_n) $$ and writing it "naturally" as a sum of $n$ terms, the $i$th term not involving subscript $i$.
Regarding the comment @MikeEarnest:
There are ways to do this that I find "unnatural", e.g. a greedy approach: throw everything into $X_1$ that fits, throw everything left into $X_2$ that fits, etc. I'm hoping there's a more symmetrical approach that would at the very least produce an equipartition, $|X_i|=|X_j|=n^{n-2}$ for all $i,j$.
To go up one more step with the examples, the below at least follows some symmetry breaking rule (greedily taking those that could be sent elsewhere from those available whose leading digit is the in the next available in a forward-cyclic manner, starting with those avoiding 1):
Below are 16 avoiding index 1.
234,243,324,342,423,432, (have to keep)
334,343,344, (could send to 2)
224,242,244, (could send to 3)
223,232,233, (could send to 4)
222 (could send to 3 or 4)
Below are 16 avoiding index 2.
134,143,314,341,413,431 (have to keep)
443,434,433, (could send to 1)
414,441,411, (could send to 3)
313,331,311, (could send to 4)
333 (could send to 1 or 4)
Below are 16 avoiding index 3.
124,142,214,241,412,421, (have to keep)
424,442,422, (could send to 1)
114,141,144, (could send to 2)
212,221,211, (could send to 4)
444 (could send to 1 or 2)
Below are 16 avoiding index 4.
123,132,213,231,312,321, (have to keep)
323,332,322, (could send to 1)
113,131,133, (could send to 2)
112,121,122, (could send to 3)
111 (could send to 2 or 3)
I have found an equipartition, but I am not inclined to call it natural.
Given $x=(x_1,\dots,x_{n-1})\in X$, find the smallest positive integer $d$ such that $x_1+d\pmod n$ does not appear in $x$. Then, letting $k=x_1+d\pmod n$, you should put $x$ into $X_k$.
To see that this is an equipartition, let $x+i$ denote the list $(x_1+i,\dots,x_{n-1}+i)$, with all entries wrapping modulo $n$, and consider these $n$ elements of $X$: $$ x,x+1,x+2,\dots,x+(n-1) $$ Each of these will be placed in a different $X_k$. Since all of $X$ can be partitioned into groups of size $n$ like above, each $X_k$ gets the same number of elements of $X$.
I would say this is not natural because it gives more importance to $x_1$ than the other entries.