Graphs with a unique transmission value

Question

If $G$ is a graph with distance function $d(x,y)$ between vertices, the transmission of a vertex $x \in v(G)$ is defined as $\sigma_{x}=\sum_{y \neq x}{d(x,y)}$. I want to know if there is a known characterization of graphs for which $\sigma_{x}$ is equal to the same number for all $x \in V(G)$.

Some easy examples: $K_{n},K_{p,p},C_{n}$. However, I found some non-regular examples as well.

UPDT: Those non-regular examples were a mistake in calculations - I still don't know if there are other, true ones.

UPDT2: As Gordon Royle pointed out, regular graphs of diameter 2 are transmission-unique; it can be easily shown that a transmission-unique graph of diameter 2 must be regular. He gave an example of diameter 3. Is there anything interesting to be said about such graphs of high diameter?

Answer 1

Here's a non-regular one:

Graph 1, order 9.
0 : 3 6 7;
1 : 4 6 8;
2 : 5 7 8;
3 : 0 6 7;
4 : 1 6 8;
5 : 2 7 8;
6 : 0 1 3 4;
7 : 0 2 3 5;
8 : 1 2 4 5;

Now the list of pairwise distances

0 2 2 1 2 2 1 1 3 
2 0 2 2 1 2 1 3 1 
2 2 0 2 2 1 3 1 1 
1 2 2 0 2 2 1 1 3 
2 1 2 2 0 2 1 3 1 
2 2 1 2 2 0 3 1 1 
1 1 3 1 1 3 0 2 2 
1 3 1 1 3 1 2 0 2 
3 1 1 3 1 1 2 2 0 

ADDED: Some more comments and another example.

Clearly one way for a regular graph to have this unique-transmission-value property is if every vertex has the same number of vertices at each distance from it. Vertex-transitive graphs, distance-regular graphs, the regular graphs of diameter 2 etc all fall into this category. You could easily fool around with graphs of higher diameter and make the girth high enough to force this to happen. These graphs are likely to be impossible to characterise any more precisely.

So the non-regular ones are perhaps more interesting. But here again, we have numerous examples due to Brendan's cartesian product example, so its not clear where to go. But it looks ugly. Here's another example, on 11 vertices, with not much obvious structure.

Graph 1, order 11.
0 : 4 7 8 9;
1 : 5 6 7 10;
2 : 5 8 9;
3 : 6 8 9;
4 : 0 7 8 9;
5 : 1 2 6 10;
6 : 1 3 5 10;
7 : 0 1 4 10;
8 : 0 2 3 4;
9 : 0 2 3 4;
10 : 1 5 6 7;

Answer 2

If I didn't get my wires crossed, the cartesian product of two transmission-regular graphs is also transmission-regular. This can be used to make examples of arbitrarily high diameter. I'll add a proof tomorrow if nobody finds a counterexample to my claim while I sleep.

ADDED: Let $G, H$ be connected graphs of order $m,n$, respectively. Then in the Cartesian product $G\times H$, we have $d_{G\times H}((u,x),(v,y)) = d_G(u,v)+d_H(x,y)$. From this it is easy to see that the transmission of $(u,x)$ is $n\sigma_u(G) + m\sigma_x(G)$. So $G\times H$ is transmission-regular iff both $G,H$ are.