How does one Segal-subdivide a 2-category? Let $\mathcal{C}$ be a small category. Then, its Segal subdivision $\text{sd }\mathcal{C}$ is a new category whose objects are morphisms of $\mathcal{C}$, and a morphism from $f:x \to y$ to $g: w \to z$ is a pair of morphisms $a: x \to w$ and $b: y \to z$ in $\mathcal{C}$ so that $f = b\circ g \circ a$. It is easy to see (by a contractible-fiber argument) that there is a homotopy equivalence of classifying spaces $B\text{Sd }\mathcal{C} \simeq B\mathcal{C}$. My first encounter with this construction was in the preprint on Morse theory and classifying spaces (link here).
Here's the question:

What is the analogue of Segal subdivision for small (strict) $2$-categories?

In particular, from a given small $2$-category $\mathcal{D}$, I would like to construct a new $2$-category $\text{sd}_2\mathcal{D}$ whose one-skeleton coincides with the ordinary Segal subdivision of the one-skeleton of $\mathcal{D}$. Some mysterious and powerful $2$-morphisms should exist, and their addition should magically make the classifying spaces of $\text{sd}_2\mathcal{D}$ and $\mathcal{D}$ homotopy-equivalent. 
What, if anything, is the precise collection of $2$-morphisms which achieve the homotopy equivalence, and where can I find this written down?
 A: Let me make a proposal. I won't check that the property you demand about nerves is satisfied. I suspect it holds, but the checking may require some work, so if I did it I would be tempted to do something with it. Since you may really want to do something with it, I leave it to you ;-)
Given a $2$-category $\mathcal C$, define the 2-category of factorizations $F(\mathcal C)$ as follows. Objects in $F(\mathcal C)$ are morphisms in $\mathcal C$, $f\colon X\rightarrow Y$. A morphism in $F(\mathcal C)$ from $f\colon X\rightarrow Y$ to $g\colon U\rightarrow V$
is a diagram in $\mathcal C$ as follows
$$\begin{array}{ccccc}
&&h\\
&X&\leftarrow&U\\
f&\downarrow&\Rightarrow&\downarrow&g\\
&Y&\rightarrow&V\\
&&k
\end{array}$$
A $2$-cell in $F(\mathcal C)$ from this morphism to
$$\begin{array}{ccccc}
&&h'\\
&X&\leftarrow&U\\
f&\downarrow&\Rightarrow&\downarrow&g\\
&Y&\rightarrow&V\\
&&k'
\end{array}$$
consists of $2$-cells in $\mathcal C$, $h\Rightarrow h'$ and $k\Rightarrow k'$ such that the pasting of
$$\begin{array}{ccccc}
&&h\\
&X&\leftarrow&U\\
f&\downarrow&\Rightarrow&\downarrow&g\\
&Y&\stackrel{k}\rightarrow&V\\
&||&\Downarrow&||\\
&Y&\stackrel{k'}\rightarrow&V
\end{array}$$
coincides with the pasting of 
$$\begin{array}{ccccc}
&X&\stackrel{h}\leftarrow&U\\
&||&\Downarrow&||\\
&X&\stackrel{h'}\leftarrow&U\\
f&\downarrow&\Rightarrow&\downarrow&g\\
&Y&\rightarrow&V\\
&&k'
\end{array}$$
Horizontal and vertical composition in $F(\mathcal C)$ are defined in the obvious way.
A: Another, somewhat similar, suggestion, in a cubical setting, which produces a double category:

In the picture I wrote the one morphisms as subdivisions, but if you want to see it as a double category, then these should be the vertical morphisms, and the top and bottom faces are the horizontal morphisms.
