The class of vertex-transitive graphs is too wild for this question to admit a coherent answer. The set of distances between vertices can vary a great deal - some examples:

- $K_n$, the complete graph on $n$ vertices. Here all vertices are distance 1 from each other;
- $K_{n,n}$, the complete bipartite graph with two lots of $n$ vertices. The set of distances is $\{1,2\}$;
- $C_n$, a cycle on $n$-vertices. The set of distances between vertices is $\{1,2,\dots, \lfloor\frac{n}{2}\rfloor\}$;

These are just a tiny set of examples from the full class - there are lots of `sporadic' examples for which the set of distances will be equally sporadic. To see this have a look at Gordon Royle's list of vertex-transitive graphs with at most 31 vertices.

**Edit**: I had a look through Gordon's list and noticed that the following is true: for every $n\leq 8$ and for every $a\leq \lfloor \frac{n}{2}\rfloor$, there is a vertex-transitive graph $X_{n,a}$ which has $n$ vertices and for which the set of distances is $\{1,\dots, a\}$. I wonder if this is true for all $n$?

**Edit 2**: The question has now been changed to ask for a proof that all vertices in a vertex-transitive graph have the same eccentricity. This is a simple consequence of vertex-transitivity, i.e. not research level.