by $m^{-1}(\sigma, \tau)=\sigma\tau$. Define a function $g_V\colon V({\mathcal{M}}(K)) \to V({\mathcal{M}}(K))$ by $g_V  := m\circ f \circ m^{-1}$. Clearly $g_V((\sigma,\tau))$ is another vertex of $V({\mathcal{M}}(K))$. Since $f$ is a simplicial isomorphism, it has an inverse $f^{-1} \colon {\mathcal{H}}(K) \to {\mathcal{H}}(K)$. It is then clear that $g_V$ has an inverse $g_V^{-1} := m^{-1} \circ f^{-1} \circ m$. Hence $g_V$ is a bijection.\
We now show that $g$ is simplicial. Consider any $\sigma  := (\sigma_1, \tau_1)(\sigma_2,\tau_2)\cdots (\sigma_k,\tau_k)\in {\mathcal{M}}(K)$. Observe that $$\begin{aligned}
g(\sigma) &= (m\circ f\circ m^{-1})(\sigma) \\
&= (m\circ f)\left( \prod_{i=1}^k \sigma_i\tau_i\right)\\
&= m\left( \prod_{i=1}^k f_V(\sigma_i) f_V(\tau_i)\right)\\
&= \prod_{i=1}^k m(f_V(\sigma_i)f_V(\tau_i)).\end{aligned}$$

Suppose that $\prod_{i=1}^k m(f_V(\sigma_i)f_V(\tau_i))= \prod_{i=1}^k (f_V(\sigma_i), f_V(\tau_i))$ as the case $\prod_{i=1}^k m(f_V(\sigma_i)f_V(\tau_i))= \prod_{i=1}^k (f_V(\tau_i), f_V(\sigma_i))$ is identical. We claim that each $(f_V(\sigma_i), f_V(\tau_i))$ is a primitive vector in ${\mathcal{M}}(K)$. Since $\sigma_i\tau_i \in {\mathcal{H}}(K)$, by the construction of ${\mathcal{H}}(K)$, we know that either $\sigma_i$ is a codimension $1$ face of $\tau_i$, or vice versa, and since $f$ is an automorphism, $f_V(\sigma_i)f_V(\tau_i) \in {\mathcal{H}}(K)$. Again, by construction of ${\mathcal{H}}(K)$, we know that either $f_V(\sigma_i)$ is a codimension $1$ face of $f_V(\tau_i)$, or vice versa. Thus, applying $m$ to $f_V(\sigma_i)f_V(\tau_i)$ will form a primitive vector $(f_V(\sigma_i)f_V(\tau_i))\in {\mathcal{M}}(K)$.\
We now must show that all primitive vectors in 