Is a graph determined up to isomorphism by its path category? This is probably easy and/or well known, but I couldn't find anything with my brain or with google.
Let $G$ be a directed graph.  Recall that the path category of $G$ is is the category whose objects are vertices of $G$ and whose morphisms are finite sequences $e_1, \ldots, e_n$ where the target of each $e_i$ is the source of each $e_{i+1}$.  (Composition of morphisms corresponds to appending sequences.)
My question is:

Is there a pair of non-isomorphic graphs with equivalent path categories?

If the answer is "no", then I have a follow-up:

Suppose two graphs have equivalent path categories.  Is every equivalence induced by an isomorphism of graphs?

 A: The path category is the free category on a graph, and this free functor is not fully faithful, since a functor between path categories need not preserve the arrows coming from the underlying graph (it is faithful but not full.) 
However, the free category functor is conservative, and even reflects equivalences into isomorphisms, as you request. The reason has to do with the structure theory of free categories. They are exactly those categories with a class $\mathcal I$ of indecomposable arrows-the word is used literally, for those arrows inexpressible as a nontrivial composition-such that every arrow is uniquely a composition of arrows from $\mathcal I$. Now, it's easy to see that indecomposability is preserved by isomorphisms of categories, so that ann isomorphism of free categories restricts to a bijection on objects and on indecomposable arrows, that is, to ann isomorphism of the underlying graphs. 
Seen another way, the free category functor is fully faithful as a functor into the subcategory of categories with functors which preserve indecomposables, and this subcategory contains all isomorphisms.
Now, we're really interested in equivalences of free categories, and equivalences need not preserve indecomposability: consider the projection of the category generated by two isomorphic objects to the terminal category. However, the isomorphism category is not free, and an equivalence of free categories actually does yield an isomorphism on underlying graphs, simply because free categories are skeletal-there are no nontrivial isomorphisms-so that every equivalence between them is actually an isomorphism!
