EDIT The question was based on an error, as it turns out. In fact my example is a category (and therefore a groupoid), by Eric Wofsey's argument. I can't remember why I thought it wasn't, and I feel a little silly, but I am very glad that it is. 


In the nerve of a small category an $n$-simplex is determined by $n+1$ objects and $n$ arrows: $$x\_0\to x\_1\to x\_2\to\dots \to x\_n
$$


What do you call a simplicial set in which an $n$-simplex is determined by a different part its $1$-skeleton, as follows:
$$
(x\_0\to x\_1\ ,\ x\_0\to x\_2\ ,\ \dots\ ,\ x\_0\to x\_n)\ ?$$
You could say that instead of a composition law for arrows you have a certain kind of decomposition law.


Have people run into this before? What is it called? Do you know any good ways of thinking about it?


I have run into something formally analogous to this (with cosimplicial commutative rings rather than simplicial sets), and I'm trying to get to know it better.


EDIT I see that I did not say all of what I meant. I should have said that given 
$$
(x\_0\to x\_1\ ,\ x\_0\to x\_2\ ,\ \dots\ ,\ x\_0\to x\_n)\ $$
(i.e. a map into $X$ from the $1$-dimensional object indicated by the above) there is a unique map $\Delta^n\to X$ extending it. By the way, the dual thing is also true in my examples: given 
$$
(x\_0\to x\_n\ ,\ x\_1\to x\_2\ ,\ \dots\ ,\ x\_{n-1}\to x\_n)\ $$
(i.e. a map into $X$ from the $1$-dimensional object indicated by the above) there is a unique map $\Delta^n\to X$ extending it. (In fact, there is also an arrow-reversing involution $X\_n\to X\_n$ for all $n$, if you know what I mean.)



This is not a category. What is it?
It's called a groupoid. Given an object $A$, call the degenerate edge from $A$ to itself the identity map at $A$. Given an edge $f:A\to B$, let $f^{-1}:B\to A$ denote the unique edge that fills in a 2-simplex whose other two edges are $f$ and the identity. Given $f:A\to B$ and $g:B\to C$, define $gf$ to be the unique edge that fills in a 2-simplex whose other two edges are $f^{-1}$ and $g$. The condition for $n=3$ using the maps $f$, $1$, and $f$ gives that $(f^{-1})^{-1}=f$. It is now easy to see that the identity maps are units for our composition operation, and $f^{-1}f=ff^{-1}=1$. 


The condition for $n=3$ now says that $(hf^{-1})(gf^{-1})^{-1}=hg^{-1}$. Setting $g=1$ tells us that $(hf^{-1})f=h$, and so we have cancellation on the right. Setting $h=1$, $f=x$, and $g=yx$ gives $x^{-1}y^{-1}=(yx)^{-1}$. Setting $f=y$, $g=z^{-1}$, and $h=xy$ now says that $x(z^{-1}y^{-1})^{-1}=(xy)z$. Since $(z^{-1}y^{-1})^{-1}=yz$, this says our composition operation is associative. It follows that composition defines a groupoid, and it is now easy to see the simplicial set is the nerve of this groupoid.


