$\operatorname{Aut}(\mathbb{H})$ acts on circles Suppose $\mathbb{H}=\{z \in \mathbb{C}\mid \operatorname{Im}(z)>0\}$.
Is it true that $\operatorname{Aut}(\mathbb{H})$ can map any circle in $\mathbb{H}$ to any circle in $\mathbb{H}$ (i.e. for any two circles $C_1,C_2$ with all their points in $\mathbb{H}$ there's some element of $\operatorname{Aut}(\mathbb{H})$ that carries $C_1$ to $C_2$)?
The only useful fact I know is that
$f\in \operatorname{Aut}(\mathbb{H}) \Rightarrow f=\frac{az+b}{cz+d}, a,b,c,d \in \mathbb{R}, ad-bc \neq 0 $.
Thank you for any help!
 A: The automorphism group here is $G=\mathrm{SL}_2\mathbb{R}$ acting by Mobius transformations (kernel $\{\pm I_2\}$).
Consider the stabilizer $\mathrm{Stab}(i)=\mathrm{SO}(2)$. Its orbits are circles symmetric about the imaginary axis intersecting it at points $\lambda i$ and $\lambda^{-1}i$ with $\lambda>1$. Thus, $K=\mathrm{SO}(2)$ fixes each of these circles. In the upper-half plane model of hyperbolic geometry (which isn't necessary for this answer, just background) $G$ acts by isometries, and these are the hyperbolic circles centered at $i$.
The Iwasawa decomposition $G=KAN$ is relevant here. That is, $K=\mathrm{SO}(2)$ has a complementary subgroup $AN$ consisting of upper triangular matrices. More precisely, $A$ consists of diagonal matrices and $N$ consists of unitriangular matrices - the letters stand for kompact, abelian, nilpotent (the explanation for these terms involves Lie theory). In other words, $G$ is a "knit product" of these subgroups, and $AN$ forms a transversal (set of coset representatives) for $G/K$.
The subgroup $AN$ is enough to turn any circle into an aforementioned one with hyperbolic center $i$. For convenience, let's call these "central" circles. You can use homotheties from $A$ to ensure a circle's max and min imaginary parts are reciprocal, then use $N$'s elements to translate it until it is symmetric about the imaginary axis. Moreover, this element of $AN$ that maps a circle to a central one is unique, and no nontrivial element of $AN$ can map one central circle to another.
Let $\Gamma$ be the set of all "central" circles. Putting the above information together, we may conclude $\Gamma$ is a complete transversal for the $G$-orbits (with $G$ acting on circles). That is, each $G$-orbit of circles may be represented by exactly one central circle from $\Gamma$. In the context of hyperbolic geometry, this means $G$ preserves hyperbolic radii and any two circles of the same hyperbolic radius are related by an element of $G$.
