I think the answer to Question 1 is yes, at least for finite graphs.
Lets $(T,E)$ be a graph. You can choose a group $G$ that is large enough to do the following: For each node $x∈V$ choose a subgroup $G_x$ (called stabilizer) such that the intersection of all subgroups is $\{1\}$ and each subgroup has a different cardinality. Then you can form the graph in the following way: $$ V= \dot\bigcup_{t∈T} \{gG_t\mid g∈G\}, \mbox{ and } F=\{(gG_s,hG_t)\mid g,h∈G, (s,t)∈E\}.$$
Then each node $t$ is broken up into $|G/G_t|$ pairwise unconnected nodes. And each edge is split up into a complete bipartite graph connecting all equivalent vertices. By changing the group size and the size of the stabilizer you can ensure that any two edges in $F$ have the same valency iff they belong to the same node in $T$.
For more information:
- Monika Zickwolff: Darstellung Symmetrischer Strukturen durch Transversale, In: Contributions to General Algebra 7, Teubner, Wien/Stuttgart
- Daniel Borchmann and Berhard Ganter: Concept Lattice Orbifolds – First Steps, In: Formal Concept Analysis, LNAI 5548, Springer, p. 22