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Here is a statement from the book:

Let G be a group of permutations that acts transitively on $V$, a set of vertices. Let $Omega$ be an orbit of $G$ on $Vtimes V$ that is not symmetric (so $Omega neq Omega^T$). Then $Omega$ is an oriented graph $G$ acts transitively on its vertices. Hence each point in $V$ has the same out-valency and the same in-valency.

I am trying to show that the latter part is true, and it's been tough. Doing any sort of matchings of arcs from $(x,y)$ to $(a,b)$ for some $x,a in V$ proves to be tough since we have a lot of $g$'s and any edge $(x,y)$ can be mapped to any edge $(a,b)$ by choosing (possibly more than one choice) a $g in G$ such that $g(x) = a, g(y) = b$.

One avenue of attack is this: Say $x,a in V$. We know that $G_x leq G$. Since the set of $g in G$ such that $g(x) = a$ is a right coset of $G_x$, we can denote $G_{x rightarrow a} = lbrace g : g(x) = a rbrace$. Now since $G$ is transitive and $Omega$ is an orbital, we know that if $(a,b) in Omega, exists g in G$ such that $(x,y)^g = (a,b)$. This $g$ is also in $G_{x rightarrow a}$. So if $langle x rangle$ are all the out-arcs of $x$, then $G_{x rightarrow a}$ applied to $langle x rangle$ will be the set of edges $langle a rangle$. So $g$ can be viewed as a permutation between $y in V : (x,y) in Omega$ and $b in V : (a,b) in Omega$. Since $g$ is a permutation, the map is 1-1. So the sets must be the same. Similar logic can be applied to obtain the same result for in-arcs.

This seems to be over-complicated for a statement in the book that is not proven. Am I missing a much simpler proof for this statement?

PS. Sorry for the excessive notation, I tried my best to be as precise and as clean as possible. Thanks!

(It is not clear from your question how you get the orientations of each edge from the automorphism group, since if there is an element of $G$ sending $(a_{1},b_{1}) mapsto (a_{2},b_{2})$ then there is also an element sending $(a_{2},b_{2}) mapsto (a_{1},b_{1})$. So I am going to assume that this orientation is defined in a way so that the action of $G$ gives an automorphism group of the oriented graph.)

It is immediate that $G$ acts transitively on $Omega$. The action of $G$ gives an automorphism group of the oriented graph. Since automorphisms must send a vertex $v$ to another vertex with the same in-degree and out-degree, the existence of a vertex-transitive automorphism group forces every vertex to have the same in- and out-degree.