Does there exist a finite nonoriented graph whose automorphism group is transitive but not generously transitive (that is, it is not true that each pair $(x,y)$ of vertices can be interchanged by some automorphism)?
I know, for example, that Cayley graphs are always generously vertex-transitive.
On
To provide a more explicit example: Take two vertices of distance 2 on the truncated tetrahedron. Observe that there is a unique path between these vertices, and that the edges in this path are distinct (only one of them is part of a triangle, for instance). So we can define a direction on this path, and thus can't hope to flip it in an automorphism.
Cayley graphs for abelian groups have generously transitive automorphism groups. In general a Cayley graph for a non-abelian group will not be generously transitive.
In particular if $G$ is not abelian and $X$ is a so-called graphical regular representation (abbreviated GRR) for $G$, then its automorphism group is not generously transitive. The key property of a GRR is that the stabilizer of a vertex is trivial.
Constructing GRRs is not trivial, but it is expected that most Cayley graphs for groups that are not abelian or generalized dicylic will be GRRs - so choosing a connection set at random usually works.