Given $n$ arrows arranged so that every arrow starts from the base of one of the arrows and ends on the base of one of the arrows, what should be meant by that two such arrangements are essentially the same? E.g. how many essentially different arrangements exists for $n=2$?
The diagram below shows only a fragment of a complete arrangement.
An arrow could also start or end on its own base.
I regard all arrangements for $n=1$ to be equivalent.
I had the feeling that there was a canonical answer, but that I just couldn't see it. Thinking about it a little more I found:
A morphism is a function between the arrows of two arrangements which in the obvious way preserves the structure, and equivalence is "isomorphic to".
The accepted answer:
a={ <1,1,1> <2,1,1>}
b={ <1,1,1> <2,1,2>}
c={ <1,1,1> <2,2,1>}
d={ <1,1,1> <2,2,2>}
e={ <1,1,2> <2,1,1>}
f={ <1,1,2> <2,1,2>}
g={ <1,1,2> <2,2,1>}
h={ <1,2,1> <2,1,1>}
i={ <1,2,1> <2,1,2>}
Obviously this nine arrangements are non isomorphic due to definition.
Each arrangement of arrows can be represented as a set of triples $ < name, head, tail> $ selected from an alphabet $\Sigma$ with the restriction that each triple has a unique name. In this representation $ \{ <0,0,0> \} $ would be your $ n = 1 $ case where an arrow starts and ends on its own base.
We then can define two arrangements to be equivalent if there is a permutation $ \Sigma \to \Sigma $ remaps one arrangement to the other.
With this definition I was able to iterate 9 examples for n=2.
{ <1,1,1> <2,1,1>} { <1,1,1> <2,1,2>} { <1,1,1> <2,2,1>} { <1,1,1> <2,2,2>} { <1,1,2> <2,1,1>} { <1,1,2> <2,1,2>} { <1,1,2> <2,2,1>} { <1,2,1> <2,1,1>} { <1,2,1> <2,1,2>}