$g(\sigma) = \prod_{i=1}^k (f_V(\sigma_i), f_V(\tau_i))$ are compatible. Suppose for the sake of contradiction they are not all compatible. The first possibility is that there are two compatible primitive vectors $(\sigma_k,\tau_k),(\sigma_{\ell}, \tau_{\ell}) \in V({\mathcal{M}}(K))$ such that

$$(f_V(\sigma_k),f_V(\tau_k))(f_V(\sigma_{\ell}),f_V(\tau_{\ell})) \not\in {\mathcal{M}}(K).$$ However, this implies that $\{f_V(\sigma_k),f_V(\tau_k)\} \cap \{f_V(\sigma_{\ell}),f_V(\tau_{\ell})\} \neq \emptyset$. Since $f_V$ is bijective, this implies that $\{\sigma_k,\tau_k\} \cap \{\sigma_{\ell}, \tau_{\ell}\} \neq \emptyset$, so that $(\sigma_k,\tau_k),(\sigma_{\ell}, \tau_{\ell})$ are not compatible, which is a contradiction. The second possibility is that $g(\sigma)$ contains a nontrivial closed $V$-path. In a similar manner to the first case, this nontrivial closed $V$-path in the image of $g$ can be pulled back to obtain a nontrivial closed $V$-path in $\sigma$, a contradiction. We conclude that $g(\sigma)\in {\mathcal{M}}(K)$ so that $g$ is a simplicial map.

Since $g_V$ is a bijection and $g$ is simplicial, it follows that $g$ is a simplicial automorphism on ${\mathcal{M}}(K)$. Define $\phi(f) := g$. We first show that $\phi$ is a homomorphism. Suppose we have $a, b \in {\mathrm{Aut}}({\mathcal{H}}(K))$. We seek to show that $\phi(a\circ b) = \phi(a)\circ \phi(b)$. By the definition of a simplicial map, it suffices to show that $\phi(a\circ b)_V = (\phi(a)\circ \phi(b))_V$. We have $$\begin{aligned}
\phi(a\circ b)_V&=& m\circ a\circ b \circ m^{-1}\\
