For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<\dots<n\}$ into the product
$\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$,
in which $1$ appears $i$ many times.
In $n$-category theory the totally ordered set with $n+1$ elements and the powerset of the set with $n$ elements admit the following categorifications called the $n$-th oriental $O(n)$ and the
lax $n$-cube $\boxtimes^n$. These are $n$-categories, whose 1-truncations are the ordered set with $n+1$ elements and the powerset of the set with $n$ elements, respectively.
Is there a $n$-functor $O(n) \to \boxtimes^n$ that sends $i$ to the increasing sequence $(0 < \dots<0<1<\dots<1)$, in which $1$ appears $i$ many times?