While thinking about free categories generated from the underlying graph of a category, I got myself a bit puzzled over whether free categories always generate new identity morphisms, distinct from the identity morphisms of the category whose underlying graph it's being generated from.
The reason I ask is because it's my understanding that by "underlying graph" what is really meant is "underlying quiver" since multiple edges with the same source and target and loops are allowed. Then it seems that the forgetful functor taking you from a category to its underlying quiver would identify identity morphisms with loops. But since the forgetful functor "forgets" facts about identity and composition, if there was an object $A$ with $f: A \rightarrow A$ such that $f \neq \mathit{id}_A$ this would be represented in the underlying graph by the vertex having two loops on it. But the underlying graph would obviously have "forgotten" which is $\mathit{id}_A$ and which is $f$.
Now, whenever I've seen an explanation of the procedure for generating a free category from a graph, it's always said that the identity morphism is given by the empty path on a vertex. I understand that if you can't assume the graph you're generating from has loops for each vertex, the empty path is the only way to guarantee every object in the free category gets an identity morphism. But suppose you're generating a free category from a quiver where every vertex has a loop, as is the case where the quiver is the underlying graph of a category. In that case, could the free category take identity morphisms to be a unique loop for each vertex in the quiver (equipped with the necessary properties with respect to composition)? Or would the identity morphisms still be the empty paths?
If it is always the case that identity morphisms of free category are the empty paths, then does it follow that the underlying graph of a free category is not the graph that generated it?
It's true that the underlying graph of a free category is never the generating graph, except in case that the generating graph was empty.
In particular, you're correct that no pre-existing loop in a graph can become the identity morphism in the free category generated by a graph. Even if ever vertex admits a loop, the graph structure admits so functorial choice of favored loops to turn into identity morphism.
However, others have had the same distaste for this phenomenon as you, and so the free category functor is sometimes defined on the category of reflexive graphs, that is to say quintuples $(V,E,s:E\to V,t:E\to V,i:V\to E| si=ti=\mathrm{id}_V)$,or more simply, graphs with a chosen loop at each vertex. This version of the free-forgetful adjunction does not generate new identity morphisms.