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.
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