Let $f_\ast\colon {\mathcal{M}}(K) \to {\mathcal{M}}(K)$ be induced by $f$. Notice that $(a,ab) \in {\mathcal{M}}(K)$, but $f_\ast((a,ab)) = (f(a), f(ab)) = (a,a) \not\in {\mathcal{M}}(K)$. In order to avoid these “degenerate" primitive vectors, we must impose the constraint that $f$ be an isomorphism.

We now show that a simplicial automorphism on a simplicial complex $K$ gives rise to a simplicial automorphism on ${\mathcal{M}}(K)$.

\[prop: induced auto\] Let $f\colon K\to K$ be a simplicial automorphism. Then the induced map $f_{\ast}\colon \mathcal{M}(K) \to \mathcal{M}(K)$ is a simplicial automorphism.

We know by Proposition \[prop: f\_\* simplicial\] that $f_*$ is simplicial. It thus suffices to show that $f_{\ast_V}\colon V(\mathcal{M}(K)) \to V(\mathcal{M}(K))$ is a bijection. Consider any $(\sigma, \tau)  \in V(\mathcal{M}(K))$. We know that $f$ is a simplicial isomorphism, so it has an inverse $g\colon K \to K$. Thus, $(g(\sigma), g(\tau)) \in V(\mathcal{M}(K))$, as an argument similar to that in Proposition \[prop: primitive to primitive\] verifies that $(g(\sigma), g(\tau))$ is primitive. Hence, $f_{\ast_V}((g(\sigma), g(\tau))) = (f\circ g(\sigma), f\circ g(\tau)) = (\sigma, \tau)$. Therefore, $f_{\ast_V}$ is surjective. Since $V(\mathcal{M}(K))$ is finite, this implies $f_{\ast_V}$ is bijective.

We can then show that the induced map respects composition.

\[lem: induced comp\] Let $f,g\colon K\to K$ be simplicial automorphisms with induced 