Let us consider the following possible exact sequence $$1 \to N_{16} \to G_{64} \to \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \to 1$$
$\bullet$ The full group is a non-Abelian finite group $G_{64}$ with 64 group elements.
$\bullet$ The normal subgroup is a non-Abelian finite group $N_{16}$ with 16 group elements.
$\bullet$ The quotient group is a Abelian finite group $Q=\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ with 4 group elements.
The rule of the games (a puzzle told by my friend):
We want to impose the following conditions:
$\bullet$ $G_{64}$ contains generators $i$, $X$ and $Y$, such that $$XY=iYX$$ with $i \cdot i=-1$.
$\bullet$ $N_{16}$ contains generators $X$ and $Y^2$. But we find that $XY^2=-Y^2X$.
$\bullet$ The quotient group is $Q=\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.
Questions: What are the names of groups $N_{16}$ and $G_{64}$? Are they product of cyclic groups with Dihedral groups?
My guess is the following
(which may not be correct, as my attempt to solve my friend's puzzle):
$\bullet$ So we imagine that the $G_{64}$ is fully generated by $i,X$ and $Y$. Here $i^4=1$. (And we may need to check that $X^4=Y^4=1$ is true(?) or not.)
$\bullet$ so $N_{16}$ can be fully generated $-1$, $X$ and $Y^2$ (?).
$\bullet$ $Q$ is generated by $i$ and $Y$ (?). The $i$ and $Y$ generators are of order 2 in $Q$, but they are of order 4 in $G$.
The question may be hard to digest in disguise, but it makes perfect sense. And I have an answer for it,
The full group $G_{64}$ contains group elements $X$, $Y$, $i$ with $$X=\begin{pmatrix}0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ \end{pmatrix}$$ $$Y=\begin{pmatrix}1&0&0&0\\ 0&i&0&0\\ 0&0&-1&0\\ 0&0&0&-i\\ \end{pmatrix}$$ where my proposal satisfies your: $$XY=iYX$$ so $$G_{64}=\langle i^{n_1} X^{n_2} Y^{n_3}|XY=iYX,n_1\in \mathbb{Z}/4\mathbb{Z},n_2\in \mathbb{Z}/4\mathbb{Z},n_3\in \mathbb{Z}/4\mathbb{Z}\rangle$$ with $i^2=-1$, $X^4=Y^4=i^4=1$, the 4-by-4 identity element.
The normal subgroup $N_{16}$ contains group elements $X$, $Y^2$, $-1$ with $$X=\begin{pmatrix}0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ \end{pmatrix}$$ $$Y^2=\begin{pmatrix}1&0&0&0\\ 0&-1&0&0\\ 0&0&1&0\\ 0&0&0&-1\\ \end{pmatrix}$$ with what you desired, $$XY^2=-Y^2X,$$ so $$N_{16}=\langle (-1)^{m_1} X^{m_2} Y^{2m_3}|XY^2=-Y^2X, m_1\in \mathbb{Z}/2\mathbb{Z},m_2\in \mathbb{Z}/4\mathbb{Z},m_3\in \mathbb{Z}/2\mathbb{Z}\rangle$$ with $i^2=-1$.
Last, we can consider the quotient group is isomorphic to $$Q \cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z},$$ in the full group $G_{64}$ the two generators of $Q$ is indeed $i$ and $Y$. They are order 4 in $G_{64}$ ($i^4=Y^4=1$), but they are order 2 in $Q$.
Everything satisfies what you imposed. And your guess is basically right. I don't have the name of the group though. Not sure how $G_{64}$ and $N_{16}$ are named.