Let $D_{2^n}$ be a dihedral group of order $2^n$. I want to find embedding of $D_{2^n}$ in $GL(2,\mathbb{Z}_{2^n})$ if such embedding exists. I know, that $D_{2^n}$ is isomorphic to the next subgroup of $GL(2,\mathbb{C})$:$$ \left< \begin{pmatrix} e^{\frac{\pi i}{2^{n-2}}} & 0 \\ 0 & e^{-\frac{\pi i}{2^{n-2}}} \end{pmatrix} , \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right> $$
But, in $\mathbb{Z}_{2^n}$ there are no elements of order $2^{n-1}$ to create similar matrices, so I'm not sure does this embedding even exist.
While $\mathbb Z/2^n\mathbb Z$ does not have an element of multiplicative order $2^{n-1}$, it has elements of additive order $2^{n-1}$, for example $2$.
One way to turn this into a $2\times 2$ matrix of order $2^{n-1}$ is to generate symmetries of $\mathbb Z/2^n\mathbb Z$ as an affine line, with generators "translation by 2" and "reflection around 0":
$$r = \begin{pmatrix}{1} & {2} \\ {0} & {1}\end{pmatrix},\ \ \ s = \begin{pmatrix}{-1} & {0} \\ {0} & {1}\end{pmatrix}.$$
It is easily checked that these are generators of a dihedral group.