$H_v$-orbits are unions of $G_v$-orbits which have cardinalities $(k+1),(k+1),k(k+1),k(k+1),\dots,k^n(k+1),k^n(k+1),\dots$. Similarly, $H_e$-orbits are unions of $G_e$-orbits whose cardinalities are $2,2k,2k,2k^2,\dots,2k^n,2k^n,\dots$ (notice here that an $H_v$-, and hence $H_e$-, orbit cannot have just $2$ elements). So the cardinality of any $H_v$-orbit has the form either $k^n+O(k^{n-1})$ or $2k^n+O(k^{n-1})$, and that of the corresponding $H_e$-orbit may have the form either $2k^n+O(k^{n-1})$ or $4k^n+O(k^{n-1})$. Hence this (common) cardinality is $2k^n+O(k^{n-1})$, which means that the radius of every $H_v$- and $H_e$-orbit is integer. Moreover, this shows that the coresponding $H_v$- and $H_e$-orbits have the same radius.

Consider now some $\tau\in H$ mapping $v$ to $e$; every $H_v$-orbit $\Omega_v$ is mapped to some $H_e$-orbit $\Omega_e$ of the same radius $r$. Notice that a dominating part $\partial\Omega_v$ of $\Omega_v$ consists of far vertices $v'$ with $d(v,v')=r$ and far edges $e'$ with $d(v,e')=r-1/2$; similarly, a dominating part $\partial \Omega_e$ of $\Omega_e$ consists of far edges $e'$ with $d(e,e')=r$. So most of such permutations $\tau$ map a good proportion of far vertces and edges (for $v$) to far edges (for $e$). See footnote for the explanation of the term `most of'.

Let $e'$ be an edge incident to $v$, and let $\Omega_{e'}$ he $H_{e'}$-orbit of radius $r$. Notice that $k^r$ of far edges for $e'$ are also far edges for $v$. So, most of them also map to far edges/vertices for $\tau(e')$, so thare are almost $k^r$ common far edges for $e$ and $\tau(e')$. This may happen only if $\tau(e')$ is a vertex incident to $e$. But $e'$ can be chosen in $k+1$ ways, while $e$ has only two endpoints. A contradiction.

$H_v$-orbits are unions of $G_v$-orbits which have cardinalities $(k+1),(k+1),k(k+1),k(k+1),\dots,k^n(k+1),k^n(k+1),\dots$. Similarly, $H_e$-orbits are unions of $G_e$-orbits whose cardinalities are $2,2k,2k,2k^2,\dots,2k^n,2k^n,\dots$ (notice here that an $H_v$-, and hence $H_e$-, orbit cannot have just $2$ elements). So the cardinality of any $H_v$-orbit $\Omega_v$ has the form either $k^n+O(k^{n-1})$ or $2k^n+O(k^{n-1})$, and that of the corresponding $H_e$-orbit $\Omega_e$ may have the form either $2k^n+O(k^{n-1})$ or $4k^n+O(k^{n-1})$. Hence this (common) cardinality is $2k^n+O(k^{n-1})$. This means that the radius of $\Omega_v$ is an integer $r$< and that of $\Omega_e$ is either $r$ or $r+1/2$.

Case 1. $\Omega_v$ and $\Omega_e$ have the same radius $r$. Consider now some $\tau\in H$ mapping $v$ to $e$; every $H_v$-orbit $\Omega_v$ is mapped to some $H_e$-orbit $\Omega_e$ of the same radius $r$. Notice that a dominating part $\partial\Omega_v$ of $\Omega_v$ consists of far vertices $v'$ with $d(v,v')=r$ and far edges $e'$ with $d(v,e')=r-1/2$; similarly, a dominating part $\partial \Omega_e$ of $\Omega_e$ consists of far edges $e'$ with $d(e,e')=r$. So most of such permutations $\tau$ map a good proportion of far vertces and edges (for $v$) to far edges (for $e$). See footnote for the explanation of the term `most of'.

Let $e'$ be an edge incident to $v$, and let $\Omega_{e'}$ he $H_{e'}$-orbit of radius $r$. Notice that $k^r$ of far edges for $e'$ are also far edges for $v$. So, most of them also map to far edges/vertices for $\tau(e')$, so thare are almost $k^r$ common far edges for $e$ and $\tau(e')$. This may happen only if $\tau(e')$ is a vertex incident to $e$. But $e'$ can be chosen in $k+1$ ways, while $e$ has only two endpoints. A contradiction.

Case 2. Assume that the radius of $\Omega_e$ is $r+1/2$. By symmetry, we may assume that $d(v,e)=r+1/2$, so that $v\in\Omega_e$. Due to the cardinalities, $\Omega_e$ should contain also either all vertices at distance $r-1/2$ from $e$, or all edges $e'$ at distance $r-1$ from $e$. In the former case, we get that an edge at distance $r-1/2$ and an edge at distance $r+1/2$ are equivalent modulo $H_v$, so $e\in\Omega_v$, which is impossible. In the latter case, we have $\sigma\in H$ with $\sigma(e')=v$, $\sigma(e)=e$. But, since $e$ lies in $\Omega_{e'}$ corresponding to $\Omega_e$, we get $e\in\Omega_v$ again.