What functors are these? The category of "just arrow" categories is equivalent to Cat as we see here.  If we have a "just arrow" category $C$, I think we can also have the set of equations, $EQ_C$, over the (compositions of) the arrows of $C$.  These equations are extra data that define $C$.  I am not sure if this is possible.  We see our "just arrow" category as a monoid of arrows with a partial composition defined over the arrows (so $D \subset C \times C$ and $f: D \rightarrow C$ defines composition).  Furthermore, the extra data that really defines $C$ are equations over compositions (words) in the arrows of $C$.  So, for every "just arrow" category $C$, there should be a set of equations that (along with the compositions) define it, and I am calling it $EQ_C$.
Consider the category of "just arrow" categories $JA$.  Next, consider two cats $A, B \in JA$ such that $EQ_A \subset EQ_B$.  I think there should be an obvious functor $F : A \rightarrow B$ which is just an inclusion of the equations in $A$ into the equations of $B$.  Is this a familiar functor? Does it have a name?
Edit: It has been pointed out that the simple criteria of $EQ_A \subset EQ_B$ does not give rise to a functor in general.  Can we define $A$ such that there does exist such a functor?  I mean, define $A$ in relation to $B$?
 A: No, such functor doesn't exist in general case. For example, let $A=\{1\}$ (the category of arrows, corresponding to the trivial monoid) and $B=\mathbb{Z}/2\mathbb{Z}$ (the category of arrows, corresponding to the multiplicative monoid of integers modulo $2$, whose arrows are $0$ and $1$). Note, that $EQ_A\subset EQ_B$, but the inclusion mapping $i\colon\{1\}\to\{0,1\}$ doesn't induce a functor.
But such inclusion functor exists if we define the notion of "just arrow subcategory". I will try to do that.
Definition. Let $A$ and $B$ be $JA$-categories (i.e. sets with partial binary operations, satisfying axioms of category). Then $A$ is a $JA$-subcategory of $B$ iff


*

*$A\subset B$;

*for every pair $(f,g)\in A\times A$, if it is composable in $B$, then it is composable in $A$ and compositions coincide;

*for every arrow $f\in A$, its left and right identities in $B$ also belong to $A$.


In this case the inclusion mapping $i\colon A\to B$ is a $JA$-functor.
The example from the first paragraph shows that $EQ_A\subset EQ_B$ doesn't imply the third axiom of this definition. I leave it as exercise to check that it also doesn't imply the second axiom.
