&=& (m \circ a \circ m^{-1}) \circ (m \circ b \circ m^{-1})\\
&=& \phi(a)_V \circ \phi(b)_V.\end{aligned}$$ By Lemma \[lem: comp\], we know that $\phi(a\circ b)_V = \phi(a)_V \circ \phi(b)_V = (\phi(a)\circ \phi(b))_V$, as desired.\
To see that $\phi$ is injective, suppose that $\phi(a) = \phi(b)$. Then we have $m \circ a \circ m^{-1} =  m\circ b \circ m^{-1}$ which implies that $a=b$.\
Finally, we show that $\phi$ is surjective. For any $g \in {\mathrm{Aut}}({\mathcal{M}}(K))$, $g$ must be induced by some $g_V: V({\mathcal{M}}(K)) \to V({\mathcal{M}}(K))$. Then construct an $f \colon {\mathcal{H}}(K)\to {\mathcal{H}}(K)$ defined by $f = m^{-1} \circ g \circ m$. We see that $\phi(f)  = m\circ (m^{-1} \circ g \circ m) \circ m^{-1} = g$, as desired. Therefore, we conclude that ${\mathrm{Aut}}({\mathcal{H}}(K)) \cong {\mathrm{Aut}}({\mathcal{M}}(K))$.

We now are able to easily compute the automorphism group of $C_n$.

\[prop: aut C\_n\] If $K = C_n$, then then ${\mathrm{Aut}}({\mathcal{M}}(C_n)) \cong {\mathrm{Aut}}(C_{2n}) $.

It suffices to show ${\mathrm{Aut}}({\mathcal{H}}(C_n)) \cong {\mathrm{Aut}}(C_{2n})$. We construct the Hasse diagram of $C_n$:

$$\begin{tikzpicture}[scale=1.7]

\node[inner sep=1pt, circle, fill=black](a1) at (0,0) {};
\node[inner sep=1pt, circle, fill=black](a2) at (1,0) {};
\node[inner sep=1pt, circle, fill=black](a3) at (2,0) {};
\node[inner sep=1pt, circle, fill=black](a4) at (4,0) {};
\node[inner sep=1pt, circle, fill=black](ak) at (5,0) {};
\node 