Consider a tree with k nodes and for each node v the vector l^{v} = (l^{v}_{0}, l^{v}_{1}, ..., l^{v}_{k1}) with l^{v}_{d} the number of leaves (!) with distance d to v. I wonder whether two nodes v, w with l^{v} = l^{w} are conjugate (I guess they are). Can anyone help me to prove this  or give a counter example?
I have a counterexample. It is not enough just to count leaves, since this doesn't take into account the number of possible ways to arrive at those leaves. Consider the graph below. A  B  C  D  E  F   G H  I I think the vector for C and D both is 002200000, since they each have two leaves at distance 2 and two leaves at distance 3. But they are not conjugate, since C has degree 3 and D has degree 2. I think this might be a minimal counterexample. 

