(Yesterday I found an answer to this exact question by mistake and today I am not being able to find it again...)
I am asked to show that no group has $D_n$ as its derived subgroup. I am also given a hint, which reads
Suppose $G$ is such that $G' = D_n$. Let $K < D_n$ be such that $K \cong \mathbb{Z}_n$ and show that $\varphi: G \to Aut(K)$, given by $\varphi(g) = c_g$, is a group homomorphism and study $\varphi(G^{(1)})$.
First of I am not sure if $\varphi(g) = c_g$ is conjugation by $g$. If it is, then I believe I proved it is indeed a group homomorphism:
$$\varphi(gh)(a) = ghah^{-1}g^{h-1} = g(hah^{-1})g^{-1} = (\varphi(g)\circ \varphi(h))(a)$$
I also know that if $\varphi$ is a homomorphism, then $\varphi(G^{(1)}) = \varphi(G)^{(1)}$ and thus we should have
$$\varphi(D_n) = \varphi(G)^{(1)}$$
and maybe the proof continues by showing that these two are actually different... Other than that, I have no clue as to how keep going.
Note that $Aut(K)$ is abelian, since $K$ is cyclic.
Thus, the kernel of $\phi : G \to Aut(K)$ must contain $G'$.
Therefore, $$\{e\} = \phi(G') = \phi(D_n)$$
therefore, conjugation by any element of $D_n$ acts as the identity on $K$. So $K \subseteq Z(D_n)$, But this is not true (for $n \ge 3$).