I came up with this question while sorting lego bricks. Is there a name for this type of problem? The so-called "Eintstein's Riddle" pops up in my mind, but I'm unable to discern if that problem is equivalent to what I'm about to describe:
Consider the set $\mathcal{S}$ whose elements are known, but their natural ordering $\mathcal{N}$ is not. $\mathcal{N}$ is something we want to find out. Then consider $\mathcal{O}$, which is a set of ordered subsets of $\mathcal{S}$. For some $\mathcal{O}$'s, we are able to infer $\mathcal{N}$ and for some we are not. Is this property of certain $\mathcal{O}$'s that can make or break the possibility of inferring $\mathcal{N}$ something that has been studied? Does it even have a name?
A concrete example: $$\mathcal{S}=\{red,blue,green,yellow\},\,\text{I secretly reveal to you that this is also $\mathcal{N}$}$$ $$\mathcal{O}_1= \{\{red,blue\},\{red,green,yellow\},\{red,blue,yellow\}\}$$ $$\mathcal{O}_2= \{\{red,blue\},\{red,green,yellow\},\{blue,green,yellow\}\}$$ From $\mathcal{O}_1$ we are unable to solve from $\mathcal{N}$ while for $\mathcal{O}_2$, this is possible. In layman's terms: $\mathcal{O}_2$, has the magic sauce property whereas $\mathcal{O}_1$ does not.
This is basically computing the transitive closure of a relation followed by a topological sort, though with a slight tweak. I don't know a specific name for it.
Here is a quick-and-dirty algorithm for seeing if your collection of ordered sets determines a unique ordering $N$ on $S$. Make a list of all pairs $(a,b)$ such that $a$ occurs before $b$ in one of your ordered subsets. Now form the transitive closure of that relation. While there are more sophisticated algorithms for this, you can simply repeatedly pass over the list, adding a new pair $(a,c)$ whenever $(a,b)$ and $(b,c)$ are already in the list and $(a,c)$ isn't. Stop when a pass turns up no new pairs.
Now to see if this transitive closure determines a unique order, look for any elements $a$ which have no predecessors. I.e., no $(b,a)$ appears in the list. If there is more than one, a unique order is not determined. Otherwise $a$ must be the first element. Now remove all pairs from the list which contain $a$, and repeat. This is basically a topological sort, but if at any step you have a choice--more than one element without predecessors--then $N$ is not uniquely determined. With the standard topological sort, you make an arbitrary choice and keep going.
If at any stage there are no elements without predecessors, that's a sign that your original relation had loops and there is no ordering on $N$ consistent with all your ordered subsets.
Alternately, once you've computed the transitive closure, just count up the number of predecessors of each element, and then sort by that number. If there is a unique ordering, then you will get $0,1,2,\ldots$, and conversely.