Connection between bijective maps and subsets of product sets in a multi-variable problem Let $X$ be a finite set. A bijective map $f: X\to X$ can be represented by a subset $A$ of $X\times X$, such that for every element $a\in X$ there is only one element $b$ so that $(a,b)\in A$, and similarly the other way around.
Now we want to generalize this, and look at subsets of multiple products of the set $X$, and considering different ways of how a subset can represent a bijective map.
The simplest new case is when we have $X^4$, and we want to get bijective maps $X^2\to X^2$. The problem to solve is to find all subsets $A\in X^4$, so that it describes bijective maps in two ways. If we denote variables as $(a,b,c,d)\in X^4$ then the subset $A$ should be such that it gives bijective maps in the directions $(a,b)\to (c,d)$ but also $(a,d)\to (c,b)$.
What is the name for such a subset and the different ways of organizing it into bijective maps? I don't really know where to look for such solutions.
The more difficult problem is when we want to go to even bigger products. A problem which came up is to look at subsets of $X^6$ and consider three ways of organizing the subset into a bijective map $X^3\to X^3$. Using variables $(a_1,a_2,a_3,b_1,b_2,b_3)$ we would want to have a bijective map in the directions $(a_1,a_2,a_3)\to (b_1,b_2,b_3)$, $(b_1,a_2,a_3)\to (a_1,b_2,b_3)$ and $(a_1,b_2,a_3)\to (b_1,a_2,b_3)$. If you wish, these are ``partial transpose'' operations, and we would want to have a bijective map in all directions.
A trivial solution is when the map $(a_1,a_2,a_3)\to (b_1,b_2,b_3)$ is the identity on $X^3$, this map satisfies also the other requirements. Are there any other solutions? Is this known, perhaps in some other disguise?
Update: If we know solution to the problem on $X^4$, then it can be used to get a trivial solution on $X^6$: the map $(a_1,a_2,a_3)\to (b_1,b_2,b_3)$ should be for example identity on the first variable, and the non-trivial bijective map on the second and third. But how to construct solutions that are not of this form?
 A: You might want to check out the concept of an orthogonal array Specifically a $t-(v,k,1)$ orthogonal array which has $k$ columns and $v^t$ rows filled with $v$ symbols in such a way that any $t$ columns contain each ordered $t$-tuple once. That article gives a $2- (4,5,1)$ example.
In your setup that is a $16$ element subset of $X^5$ with $X=\{1,2,3,4\}$ so that you can read it $120$ ways to get a bijection from $X^2$ to $X^2$ such as by interpreting $(a,b,c,d,e)$ as $(e,c)\rightarrow  (a,d)$.
LATER
Here are a couple of solutions for $X^6$ , The first might satisfy you or you might feel it is just your identity example in disguise:
Take three permutations $\alpha,\beta,\gamma$ of $X$ and then these sextuples $$(u,v,w,\alpha u,\beta v,  \gamma w)$$
This gives what you want and one more: 
Using variables $(a_1,a_2,a_3,b_1,b_2,b_3)$ we would have a bijective map in the directions $(a_1,a_2,a_3)\to (b_1,b_2,b_3)$, $(b_1,a_2,a_3)\to (a_1,b_2,b_3)$ and $(a_1,b_2,a_3)\to (b_1,a_2,b_3)$. And also $(a_1,a_2,b_3)\to (b_1,b_2,a_3)$.
In the case that $\alpha,\beta$ and $\gamma$ are the identity map, all four of these are the identity map on $X^3.$

If you take odd $n$ and the sextuples $$(u,v,w,u+v,v+w,w+u) \mod n$$ for $X=\{0,1,\dots,n-1\}.$ You get less than a full orthogonal array but $17$ of the $20$ ways to pick three positions work, including the ones you want.
This is essentially multiplying vectors $\left( \begin{array}{c} u\\ v\\ w \end{array} \right)$in $X^3$ by the matrix $$M=\left(\begin{array}{cccccc} 1&0&0\\ 0&1&0 \\0&0&1\\ 1&1&0\\ 0&1&1\\ 1&0&1 \end{array}\right)$$
The choices which work correspond to the triples of rows giving a matrix with an inverse $\mod n.$ That should generalize.
