Algebras with the same Auslander -Reiten quiver have isomorphic categories of modules?

Maybe this is a stupid question but suppose you have two quivers $Q$ and $Q'$ (maybe we can suppose some nice property, like not having oriented cycles, so that kQ has finite dimension over $k$) whose Auslander Reiten quivers $A$ and $B$ are isomorphic as quivers in the sense of graph theory. Maybe we can also suppose that the isomorphism preserves the Auslander Reiten translation $\tau$ as well. Does that imply that the category of representations of $Q$ is "the same" as the category of representations of $Q'$, $Rep_Q \cong Rep_{Q'}$? Or to put it in algebraic terms, $mod_{kQ} \cong mod_{kQ'}$ ?

I find this question baffling because it looks natural but I havent been able to find any reference to it and it doesn't seem obvious. Simplify the problem and assume you want to show that $ind_{kQ} \cong ind_{kQ'}$, where $ind$ means the category of indescomposable representations and morphisms between indescomposable representations. An odd category, I know (very non abelian). To find a functor that is an equivalence between the two categories I need to define it over any morphism between two indescomposable representations. The problem is that, as far as I know, not every morphism is a composition of irreducible morphisms, and thus using the isomorphism between the two Auslander-Reiten quivers is more difficult.

What is valid is that any morphism $f:M \longrightarrow N$ that is not an iso, can be written as a sum of compositions of irreducible morphisms but only if $Rad^n(-,N) =0$ or $Rad^n(M,-)=0$ for some $n$. Which is something that doesn't hold in general.

Since you assume that you are given $\tau$ as part of the data, you can identify the indecomposable projective modules, since they are the ones on which $\tau$ vanishes. But the AR quiver restricted to the projective modules is the opposite of $Q$. So you can recover $Q$.
• Yes to the first point. In total generality you can't recognize the projectives without more, though: consider $A_3$ with linear orientation. Or affine $A_3$ with one source, one sink, non-adjacent. Though in both of these examples, the point somehow is that there is a different subquiver which looks exactly like the projectives. So if you took it for the projectives by mistake, you would still get the right $Q$! But I am not sure what happens in general. – Hugh Thomas Nov 11 '17 at 5:24