Let $G$ be a k-regular bipartite graph with $k > 0$. Then it is known that the two sets which partition the vertex set of $G$ have the same cardinality.
However I am interested in connected $k$-regular bipartite graphs which show an even deeper "symmetry", namely that every node is "indistinguishable" from all the other, in the sense that the graph should "look the same wherever you look at it" (I do not how to formalize this, though). The examples I have in mind are the toric square grid (4-regular) or the hexagonal toric grid (3-regular). Being embeddable on a toric surface seems helpful to fullfil the property I am asking, but it could very well be not necessary.
Here are my questions.
1) For every $k >0$, how many such graphs exists?
2) If there exists a $k$-regular bipartite graph with such property, is it unique?
3) If not, is there a way of classifying/listing/constructing them all for fixed $k$?
4) A connected $2$-regular bipartite graph is always a bipartite chain?
I guess that questions like these are closely related to uniform tilings of the plane/torus, but since I am not familiar with such topics, I could be wrong.
It is not clear from your question whether you want your graphs to be vertex transitive, or merely have the action transitive on each colour class. Nor is it clear whether you are interested in infinite graphs.
I consider the finite case. If $k=1$ we have only $K_2$. If $k=2$ we have only cycles of even length. If $k\ge3$ however, there are infinitely many finite vertex-transitive bipartite graphs with valency $k$, and there is no hope of classifying them. Just to give one class, if we choose three involutions at random for the symmetric group, then the corresponding Cayley graph is cubic and vertex-transitive (and connected with probability one). If the graph is not bipartite, its direct product with $K_2$ is still cubic and vertex transitive, and is bipartite as well.
Note that bipartiteness is no real restriction, because of the "direct product with $K_2$" trick.
Weakening the requirement from vertex transitive to transitive on colour classes just allows more examples.