You don't say how you are thinking about exchange graphs, but I am going to interpret this as a question about cluster algebras. (If it is a question about cluster categories, then the final answers are the same, but there is more preliminary background to explain why the answers are the same as for cluster algebras.)
If the underlying graph is a tree (in particular, any Dynkin type), it is because we can change any orientation of a tree into any other orientation by source-sink moves, which correspond to very simple mutations. Proof: Induction on the number of vertices. Let $T$ be a tree, let $u$ be a vertex of degree $1$, let $v$ be the unique neighbor of $u$ and let $T'$ be the tree where we delete $u$. Take two orientations $\mathcal{O}_1$, $\mathcal{O}_2$ of $T$ and restrict them to $T'$. By induction, we can change from $\mathcal{O}_1|_{T'}$ to $\mathcal{O}_2|_{T'}$ by source-sink moves in $T'$. Apply those same moves to $\mathcal{O}_1$. We have the problem that $v$ might be a source or sink for $\mathcal{O}|_{T'}$, but not in $\mathcal{O}$, but then we can always fix that by applying a source-sink move at $u$ immediately before $v$. When we are done, we will either have gotten to $\mathcal{O}_2$ or to an orientation which differs from this one by reversing the edge $(u,v)$; if the latter, apply one more sink-source move at $u$. $\square$.
If your quiver has cycles, this isn't true. A cyclic orientation of the $n$-cycle is mutation equivalent to $D_n$, and hence of finite type, but an acyclic orientation of the $n$-cycle is of infinite type.
I also don't think this is true if we restrict to acyclic orientations. Consider the $4$-cycle, where in one orientation we have two clockwise and two counter-clockwise edges, and in the other we have three and one. The corresponding cluster complexes definitely are not isomorphic -- the first has $4$ vertices which are in infinitely many clusters, and the latter has $3$. (This can be seen from the combinatorial model using triangulations of the annulus; the former case corresponds to triangulations with $2$ vertices on the inner ring and $2$ on the outer ring, the latter has $3$ and $1$. I should also be able to translate it into real Schur roots if you need.) I don't immediately see a proof that the exchange graphs are nonisomorphic, but they have no reason to be isomorphic.
