Let $Q$ and $R$ be two acyclic quivers which differ only in the directions of their arrows (i. e., the underlying undirected graphs are the same).


**1.** Does there exist an isomorphism of additive categories $\mathrm{Rep}Q\to\mathrm{Rep}R$ ?


At the moment, I am not even 100% sure about the weaker statement that there exists an equivalence of additive categories $\mathrm{Rep}Q\to\mathrm{Rep}R$. (My intuition for this is pretty much zero.)


In before reflection functors - they are not isomorphisms. (But yes, it's a nice exercise to see that we can obtain $R$ from $Q$ by a sequence of admissible reflections, where an "admissible reflection" means choosing some sink in $Q$ and switching all the arrows to $Q$. Unfortunately, the corresponding reflection functors may turn some nonzero representations of $Q$ to zero.)


**2.** Are the representation rings of $Q$ and $R$ isomorphic? The isomorphy of their additive groups follows from Gabriel's theorem, but it is not clear to me what this does to tensor products.


**1+2.** Does there exist an isomorphism of tensor categories $\mathrm{Rep}Q\to\mathrm{Rep}R$ ?


**3.** If some of these answers are No, what if we restrict ourselves to Dynkin quivers?


I am new to quivers, so I'm sorry if this has been already talked over a dozen of times here.



Acyclic quivers differing only in arrow directions: functorial isomorphism of representation categories?
The categories are not equivalent. In fact, an acyclic quiver is determined (up to non-unique isomorphism) by the equivalence class of its category of representations. The construction is simple: Find the isomorphism classes of simple objects. These are in bijection with the vertices. For any two simple objects $S$ and $S'$, the number of arrows from vertex $S$ to vertex $S'$ is the dimension of $\mathrm{Ext}^1 (S, S')$. The nonuniqueness is because we need to choose a basis of $\mathrm{Ext}^1 (S, S')$.




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There is a lot more to say on this subject, but it doesn't go in the direction your questions are pointing. When the underlying graphs of $Q$ and $R$ are trees, then it is true that the bounded **derived** categories of $\mathrm{Rep} \ Q$ and $\mathrm{Rep} \ R$ are isomorphic. The basic point here is that, although the reflection functors are not equivalences of categories, the derived reflection functors are equivalences of derived categories. Of course, every Dynkin diagram is a tree, so in particular this is true for Dynkin diagrams.


There is a version of this for non-trees, but I don't know a reference for it nor the exact statement and I don't want to hunt one down until I know whether derived categories are something that you love or that you fear. The proof, of course, is completely different.


Here is a result of Kac you might be happier with. Let our ground field be a finite field $\mathbb{F}\_q$ and fix a dimension vector $d$. Then the number of isomorphism classes of representations of $Q$ and $R$, of dimension $d$, over $\mathbb{F}\_q$, is the same. ([Infinite root systems, representations of graphs and invariant theory](http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=171997), Theorem 1.) 


Morally, one wants to work over an arbitrary field. The statement then is a statement about the stacks of $Q$ and $R$ representations of dimension $d$. But, again, to formulate this you need to know about a lot of machinery: stacks, perverse sheaves, derived categories again.




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In short, the categories are not equivalent. There are very close relations between them, but the best formulations of those relations use sophisticated category theory. You can often see shadows of these relations by counting points over $\mathbb{F}\_q$.


