**Yes, though I don’t know a clean description.**  Several relevant constructions are discussed in §8 of Aitchison 1986/2010, [*The geometry of iterated cubes*](https://arxiv.org/abs/1008.1714), especially §8.1–8.3; the following description is from my notes when reading that paper a few years back, and I think it was a working-out from the constructions there, but I can’t quite see now how it comes from them.

I’ll describe the map in terms of parity-complexes — each cell of the oriented $n$-simplex is sent to a set of cells of the lax $n$-cube. I follow Aitchison’s notation: cells of the $n$-simplex are subsets of $[n]$, written without punctuation as e.g. $014 \subseteq [5]$; cells of the $n$-cube are words of length $n$ from $\{{-},0,{+}\}$.

The general case is as follows: Given a cell $\sigma$ in the $n$-simplex, read it as splitting the $n$ positions of a potential cube cell into blocks, with each $i \in \sigma$ meaning “split after the $i$th position”: so  $125 \subseteq [5]$ divides $\newcommand{\w}[1]{(#1\,)} \newcommand{\u}{\,\_}\newcommand{\s}{\,/} \w{\u\u\u\u\u}$ into $\w{\u \s \u \s \u \u \u \s}$.  (Note the first+last blocks may be empty, but interior blocks never are.)  Now take $F(\sigma)$ to be the set of all words of the following form: their first block (under the splitting specified by $\sigma$) must be all $\newcommand{\p}{\mathord{+}}\newcommand{\m}{\mathord{-}} +$; their last block must be all $-$; and each interior block must be of the form $\m^i\;\!0\;\!\p^j$.  So in the example above, $F(125) = \{\p0\m\m0,\p0\m0\p,\p00\p\p\}$.

Some examples, in the case $n=3$ (of course, drawing some pictures is recommended here):
$$\begin{align*}
F(0) &= \{\m\m\m\} \\
F(1) &= \{\p\m\m\} \\
F(2) &= \{\p\p\m\} \\
F(01) &= \{0\m\m\} \\
F(12) &= \{\p0\m\} \\
F(02) &= \{\m0\m,0\p\m \} \\
F(03) &= \{\m\m0,\m0\p,0\p\p \} \\
F(012) &= \{00- \} \\
F(013) &= \{0\m0,00\p \} \\
\end{align*}$$

So $F$ gives a map from simplex cells to sets of cube cells.  $F$ respects dimension: the splitting induced by a $k$-cell $\sigma$ has $k$ interior blocks, so each cube cell in $F\sigma$ contains $k$ zeroes, i.e. is a $k$-cell.  And the sets $F\sigma$ are pairwise disjoint, and satisfy the condition $s(F\sigma) \sqcup F(t \sigma) = F(s \sigma) \sqcup t(F \sigma)$.  They thus induce the desired map of strict $n$-categories from the oriented $n$-simplex to the lax $n$-cube.

I’m sure there must be a better way to present these — perhaps e.g. following abstractly from the presentation of the oriented simplices as iterated coning, compared to the presentation of the lax cubes as iterated lax product with the interval — but I haven’t seen it given anywhere.

(Incidentally, I strongly encourage avoiding the term *orientals* — outside maths, its main meaning as a noun in English is as a rather dated racist slur.  By the standards of the 1980s, when Street introduced the term mathematically, it was a piece of mildly questionable-taste wordplay; today I think many more people would agree on finding it pretty distasteful.)