TITLE: What are the endofunctors on the simplex category?
QUESTION [14 upvotes]: Is there a 'classification' of the endofunctors F: Δ --> Δ where Δ denotes the simplex category with objects [n] and the weakly monotone maps [m] --> [n] as morphisms (Actually, I don't know if 'simplex category' is the right name)?
For instance there is a shift functor S: Δ --> Δ defined by S([n])=[n+1] on objects and S(d): [m+1] --> [n+1] being d on [m] and mapping m+1 to n+1 for a morphism d: [m]-->[n]. Hence for a simplicial set X one gets a path-object XoS.

REPLY [17 votes]: To carry Charles' train of thought further:
By an 'interval' let us mean a finite ordered set with at least two elements; let $Int$ be the category of intervals, where a morphism is a monotone map preserving both endpoints.
The simplicial set $\Delta^1$ can be viewed as a simplicial interval. That is, this functor $\Delta^{op}\rightarrow Set$ factors through the forgetful functor from $Int$ to $Set$. In fact, the resulting functor $\Delta^{op}\rightarrow Int$ is an equivalence of categories. 
This extra structure (ordering and endpoints) on $\Delta^1$ is inherited by Charles' $K_1=F^*\Delta^1$; it, too, is a simplicial interval.
There aren't that many things that a simplicial interval can be. Its realization must be a compact polytope with a linear order relation that is closed. That makes it at most one-dimensional, and makes each component of it either a point or a closed interval. Simplicially each of these components can be either a $0$-simplex, or a $1$-simplex with its vertices ordered one way, or a $1$-simplex with its vertices ordered the other way, or two or more $1$-simplices each ordered one way or the other and stuck together end to end. 
The three simplest things that a simplicial interval can be are: two points, a forward $\Delta^1$, and a backward $\Delta^1$. These arise as $F^*\Delta^1$ for three examples of functors $F:\Delta\rightarrow \Delta$, the only examples that satisfy $F([0])=[0]$, namely the constant functor $[0]$, the identity, and "op".
It's clear that any functor with $F([0])=[n]$ has the form $F_0\coprod\dots\coprod F_n$ where $F_i[0]=[0]$ for each $i$. This means that the corresponding simplicial interval can be made by sticking together those which correspond to the $F_i$. For example, the 'shift' functor mentioned in the question is $id\coprod [0]$; Reid mentioned $id\coprod id$ and $op\coprod id$. These correspond respectively to: a $1$-simplex with an extra point on the right, two $1$-simplices end to end, and two $1$-simplices end to end one of which is backward. As another example, the constant functor $[n]$ corresponds to $n+1$ copies of (two points) stuck together end to end, or $n+2$ points.
In short, every functor $F:\Delta\rightarrow \Delta$ is a concatenation of one or more copies of $[0]$, id, and op. I can more or less see how to prove this directly (without toposes or ordered compact polyhedra).