Take a 9-cycle with vertices $\{0,\ldots,8\}$ and join new vertices of degree one to 0, 3, 6. Delete the vertex 8 from the cycle, producing a tree $T$ on 11 vertices. Then $T\backslash 2\cong T\backslash 5$, but no automorphism of $T$ maps 2 to 5.

Now there is a theorem that if $u$ and $v$ are vertices in graph $X$ and $X\backslash u$ and $X\backslash v$ are cospectral and their complements are cospectral, the generating functions for closed walks at $u$ and for all walks at $u$ are equal to the corresponding generating functions for walks at $v$. (This is a consequence of results from chapter 4 in my book "Algebraic Combinatorics".)

So vertices 5 and 8 cannot be distinguished by your criterion. (My example is quite likely not minimal.)

Take a 9-cycle with vertices $\{0,\ldots,8\}$ and join new vertices of degree one to 0, 3, 6. Delete the vertex 8 from the cycle, producing a tree $T$ on 11 vertices. Then $T\backslash 2\cong T\backslash 5$, but no automorphism of $T$ maps 2 to 5.

Now there is a theorem that if $u$ and $v$ are vertices in graph $X$ and $X\backslash u$ and $X\backslash v$ are cospectral and their complements are cospectral, the generating functions for closed walks at $u$ and for all walks at $u$ are equal to the corresponding generating functions for walks at $v$. (This is a consequence of results from chapter 4 in my book "Algebraic Combinatorics".)

So vertices 2 and 5 cannot be distinguished by your criterion. (My example is quite likely not minimal.)

Take a 9-cycle with vertices $\{0,\ldots,8\}$ and join new vertices of degree one to 0, 3, 6. Delete the vertex 8 from the cycle, producing a tree $T$ on 11 vertices. Then $T\backslash 5\cong T\backslash 8$, but no automorphism of $T$ maps 5 to 8.

Now there is a theorem that if $u$ and $v$ are vertices in graph $X$ and $X\backslash u$ and $X\backslash v$ are cospectral and their complements are cospectral, the generating functions for closed walks at $u$ and for all walks at $u$ are equal to the corresponding generating functions for walks at $v$. (This is a consequence of results from chapter 4 in my book "Algebraic Combinatorics".)

So vertices 5 and 8 cannot be distinguished by your criterion. (My example is quite likely not minimal.)

Take a 9-cycle with vertices $\{0,\ldots,8\}$ and join new vertices of degree one to 0, 3, 6. Delete the vertex 8 from the cycle, producing a tree $T$ on 11 vertices. Then $T\backslash 2\cong T\backslash 5$, but no automorphism of $T$ maps 2 to 5.

Now there is a theorem that if $u$ and $v$ are vertices in graph $X$ and $X\backslash u$ and $X\backslash v$ are cospectral and their complements are cospectral, the generating functions for closed walks at $u$ and for all walks at $u$ are equal to the corresponding generating functions for walks at $v$. (This is a consequence of results from chapter 4 in my book "Algebraic Combinatorics".)

So vertices 5 and 8 cannot be distinguished by your criterion. (My example is quite likely not minimal.)