This question is probably a little confused. I'm teaching myself some geometry of the hyperbolic plane, and this statement from the Wikipedia page on the Poincaré half-plane model is tripping me up:
The group of all isometries of H, sometimes denoted as Isom(H), is isomorphic to $\rm {PS^*L}(2,\Bbb R)$. This includes both the orientation preserving and the orientation-reversing isometries. The orientation-reversing map (the mirror map) is $z\mapsto-\overline z$.
This seems to imply that $z\mapsto-\overline z$ is represented by an element of $\rm {PS^*L}(2,\Bbb R)$, but I can't find real numbers $a,b,c,d$ such that $\frac{az+b}{cz+d}=-\overline z$. What am I missing?
Maybe in this context we're supposed to view $\Bbb H$ as the quotient of $\Bbb C$ by the relation $z\sim\overline z$, so the mirror map is really just $z\mapsto-z$, represented by $(\begin{smallmatrix} -1 & 0\\ 0 & 1 \end{smallmatrix})\in \rm {PS^*L}(2,\Bbb R)$, but it was written as $z\mapsto-\overline z$ to show how it acts on the upper half plane. Is that right?
Yes, your idea is essentially the correct interpretation. The group $PS^*\!L(2,\Bbb R)$ is the set of Möbius transformations that map $\mathbb{R}\cup\{\infty\}$ to itself. Not all of these map the upper half-plane to itself: some of them (namely, those with negative determinant) map the upper half-plane to the lower half-plane. To get an isomorphism between $PS^*\!L(2,\Bbb R)$ and $\operatorname{Isom}(\mathbb{H})$, these Möbius transformations are turned into isometries of the upper half-plane by composing them with complex conjugation.
So, the reflection $z\mapsto-\overline{z}$ is not a Möbius transformation, but instead corresponds to the Möbius transformation $z\mapsto -z$ which sends the upper half-plane to the lower half-plane.