A graph with distnace regularity and degree $\ge \sqrt(n)$ Following may be a very silly question, and the answer may be trivial but none the less I am unable to resolve it one way or another.
Given a connected $d$-regular graph $G$ with $n$ vertices and $d \ge \sqrt(n)$ having the following property:


*

*The number of vertices at a fixed distance $h$ way from some vertex $u$ does not depend on the vertex and this is true for any $1 \le h \le diam(G)$. Where $diam(G)$ is the diameter of $G$.


Then, $diam(G) = 2$.
 A: This claim is not true. Consider the following example:

For any $x \in \{a,b,c,d\}$, we have a cycle $x_1 - x_2 - x_3 - x_4 - x_1$ and for any $i \in \{1,2,3,4\}$ we have a cycle $a_i - b_i - c_i - d_i - a_i$; these are the only edges in the graph. It is easy to check that this is $4$-regular, connected, and $n = 16$. To check whether the special property holds, note that there exist graph isomorphisms that can swap any two letters, or any two numbers. Composing these shows that for any $x,y \in \{a,b,c,d\}$ and $i,j \in \{1,2,3,4\}$, there is an isomorphism that sends $x_i$ to $y_j$. Their $h$-neighborhoods must also then be associated under this isomorphism for any $h$; in particular, they are the same size.
But there is no path $a_1 \leadsto c_3$ of length less than four. This can be verified by noting that every edge fixes the letter and increments/decrements the number by one or vice versa (we assume that the labels wrap around, e.g. incrementing $d$ gives you $a$ and so on). It requires two edges to go from $a$ to $c$, and two edges to go from $1$ to $3$. Thus, $diam(G) \geq 4$ (in fact, we actually have equality here). It is not difficult to generalize this to work for any even degree $d = 2k$. The idea will be to take $d$ distinct $d$-cliques, remove the "main" diagonals from each one, and then construct $d$-cycles which intersect each clique exactly once in the same order (in the same manner as above); the resulting graph should have diameter $2 + d/2$.
