I have a hexagon with edges $A,B,C,D,E,F$ and its symmetry group $D_6$.
I want to prove that $D_6 = H \rtimes M$ given the subgroups $H = \{ g \in D_6 \,\,|\,\, g \text{ permutes } \{A,C,E\} \}$ and $M = \{1, s\}$.
The definition of the semi-direct product is that the following holds: $(h_1,m_1) \cdot (h_2,m_2) = (h_1h_2, \,m_1 \cdot \alpha(h_1)(m_2))$,
where $\alpha$ is a map from $H \rightarrow \text{Aut}(M)$. Since $M \simeq \mathbb{Z}_2$, the only element in $\text{Aut}(M)$ seems to be the identity, so the semi-direct product reduces to the direct product.
What am I missing?
Your observation is correct: the automorphism group of $M$ is trivial. It might be that you're using a different definition for the semi-direct product than you should. The direction of the operator suggests that $M$ acts on $H$, so you should be looking at automorphisms of $H$, not $M$. Often one uses the symbol $\rhd$ denotes action, so $A\rhd B$ means $A$ acts on $B$; likewise, $A\lhd B$ means that $B$ acts on $A$. Thus the semi-direct product $A\rtimes B$ means that $B$ is acting on $A$ (coupled with the Cartesian product notation).
Edit: See this PDF. It works a very similar problem out.