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

Edit: I forgot to mention that each orbit in $(V,F)$ must be connected in certain cases (e.g. considering the lattice family $M_n$ as graphs). For each node $t∈T$ choose a (connected) Cayley graph $C_t=(G/G_t,E_t)$ and add its edges to $F$ so that the above formula reads $$F=\{(gG_s,hG_t)\mid g,h∈G, (s,t)∈E\}∪⋃_{t∈T}E_t.$$

Addendum: Size estimation.

Assuming the node are numbered by $1,…,n$, one way to assure that each node has different cardinality is to split each node $i$ into the circle $C_{(n+3)k^i}$. So we get an exponential upper bound.

However, in many cases a much smaller size can be achieved as you have to split up only as many nodes as a minimal system of generators of the automorphism group has (split up nodes instead of linking them to copies in Daniels Answer). E.g., for the complete graph $K_n$ the nodes can be replaced by circles of length $4$ to $n+4$ leading to a total size of $\frac{n(n+1)}{2}+3n$ nodes, which is polynomial. This is probably not the lower bound as it should be possible to replace certain circles by smaller Cayley graphs.