Quotient by action of this matrix group is $S^2$? Let $M^{2 \times 2}_{\neq 0}$ denote the set of all non-zero $2 \times 2$ real matrices. Note that this set contains non-invertible elements. By an orthogonal matrix, let's mean a matrix $A$ such that $AA^\top$ is a non-zero scalar multiple of the identity. Note this is different to the usual definition. Write $\mathrm{Orth}^{2 \times 2}$ for the group of all real orthogonal $2 \times 2$ matrices. Let $\mathrm{Orth}^{2 \times 2}$ act on $M^{2 \times 2}_{\neq 0}$ by left-multiplication. I have a hunch that the resulting quotient space is homeomorphic to $S^2$. Is this correct? If so, what's the homeomorphism?
Remark. It's explained in the comments that this is the same as trying to find $$\mathbb{RP}^3/O(2).$$ I actually think trying to find $\mathbb{RP}^3/SO(2)$ is the better question, and I think an assumption of orientation-preservingness was implicit in my thinking. However I am interested in both questions.
 A: Steve D's comment is correct. 
First, your action is the linear action of $O(2)$ on the sum of 2 of its defining representations, $\Bbb R^2 \oplus \Bbb R^2$. 
Now first, observe that if we write $\Bbb R^2 \oplus \Bbb R^2 = \langle 1, i\rangle \oplus \langle j, k\rangle$, then $SO(2)$ acts on each of these in the standard way, and in fact this gives us the inclusion $SO(2) \hookrightarrow S^3$, thinking of $SO(2)$ as the unit complex numbers. The quotient is $S^2$, and the projection map $S^3 \to S^2$ is the Hopf fibration. 
What remains is to identify both your $\pm 1$ action on the leftover action of $O(2)/SO(2)$. As Steve D writes in the comments, the $\pm 1$ action on $S^2$ is trivial, as it lives inside $SO(2)$ to begin with. 
Write $A$ for the diagonal matrix $\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$. It will be helpful to write elements of $S^2$ as elements of $\Bbb{CP}^1$. The action given by this matrix on each copy of $\Bbb C$ is complex conjugation; so we are looking at the action $[z:w] \mapsto [\overline z: \overline w]$. This has fixed points equal to $\Bbb{RP}^1$ and swaps the two hemispheres of $S^2$. 
Therefore, the quotient of your action is $D^2$. It might be helpful to observe that we may identify this with $[z, 1] \subset \Bbb{CP}^1$ where $\|z\| \leq 1$. Correspondingly, it seems that a near-global slice for your action would be to take upper-triangular matrices with top left term 1 and right column $\|z\| < 1$. ($O(2)$ acts on those with $\|z\| = 1$.) 
It is perhaps interesting that this $SO(2)$ action on $S^3$ actually does lift to a free action of a $\Bbb Z/2$-extesion of $SO(2)$. The group is $\text{Pin}(2) = S^1 \cup j S^1 \subset S^3$. Instead of acting by complex conjugation on each summand, $j$ swaps the two summands and does a 90 degree rotation (so $j^2 = -1$). 
