A well-quasi-ordering is a reflexive and transitive relation $\preceq$ such for any infinite sequence of the form $x_1,x_2,x_3,x_4,\dots$, there is $i,j$ such that $x_i \preceq j_i$. A well-founded relation $R$ has the property that for any infinite sequence of the form $x_1,x_2,x_3,x_4,\dots$, there is $i$ such that $\lnot x_i R x_{i+1}$.
Is there some way we can associate a well-quasi-ordering with a given well-founded relation? Its domain does not necessarily need to be on the same set as the well-founded relation.
We might be tempted to simply used $\lnot R$, but this may not be reflexive or transitive (indeed, it won't be if $R$ is reflexive). We might take its closure over reflection and completion, but then it might still not be a well-quasi-order.
(Note that we can easily get from a $\preceq$ to $R$. Simply take the set of finite sequences such that there is no $i,j$ with $x_i \preceq x_j$, and the relation "is the beginning of". Perhaps this operation will have some sort of adjoint functor.)
Ever well-founded relation has a genuine well-ordering attached to it, namely its rank. We can view this as a well-ordering of subsets of the domain of the relation, namely those sets of elements of a given rank.
Here, rank is defined as follows:
The rank of the whole well-founded relation $R$ on $X$ is the least ordinal $\alpha$ such that there is a map $f:X\rightarrow \alpha$ such that for all $a,b\in X$ we have $aRb\implies $f(a)>f(b)$.
The rank of an element $a\in X$ with respect to $R$ is the smallest value of $f(a)$, as $f$ ranges over the maps satisfying the condition above.
In some sense this is trivial, but it does seem to me the only natural construction. Meanwhile, if we require the domain to stay the same to rule out this sort of thing, it seems clear that there won't be one: what well-quasi-ordering on $X$ do you associate to the empty relation on a set $X$?