Assume that $n \geq 3$. Is there a subgroup $H \leq {\rm GL}_{2}(\mathbb{Z}/2^{n} \mathbb{Z})$ whose order is a power of $2$ so that
$\bullet$ $\det : H \to (\mathbb{Z}/2^{n} \mathbb{Z})^{\times}$ is surjective, and
$\bullet$ $H/\Phi(H) \cong (\mathbb{Z}/2\mathbb{Z})^{2}$, and
$\bullet$ $H$ contains $-I = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$, and
$\bullet$ $H$ contains all matrices of the form $I + 2^{n-1} X$ for $X \in M_{2}(\mathbb{F}_{2})$?
Motivation: Suppose that $E$ is an elliptic curve defined over a number field $K$, $E(K)$ contains a point of order $2$, and $K \cap \mathbb{Q}(\zeta_{8}) = \mathbb{Q}$. What is the smallest number of quadratic subextensions of $K(E[2^{n}])/K$? The smallest possible is $3$, namely $K(i)$, $K(\sqrt{2})$ and $K(\sqrt{-2})$. The elliptic curves for which this occurs have the property that for any positive integer $n$, the image of the mod $2^{n}$ Galois representation is a subgroup of ${\rm GL}_{2}(\mathbb{Z}/2^{n} \mathbb{Z})$ that satisfies the first and second conditions above. If $E/K$ is non-CM, there is a positive integer $n$ for which this subgroup contains all matrices $\equiv I \pmod{2^{n-1}}$. For more detailed information and context, see the paper here.
Is there ever a case where $E$ and all of its quadratic twists have this property? This would occur if $-I$ is in the image of the mod $2^{n}$ Galois representation attached to $E$, which is the motivation for the fourth condition.
Note: I have found all subgroups satisfying the first three conditions for $3 \leq n \leq 9$ and none satisfy the fourth condition. I noticed that for $5 \leq n \leq 9$, if $H \leq {\rm GL}_{2}(\mathbb{Z}/2^{n} \mathbb{Z})$ satisfies the first three conditions, then the maximal subgroup $$ \{ g \in H : \det(g) \equiv \pm 1 \pmod{8} \} $$ is abelian, which forces $H$ to be rather small.
