Do conformal mappings other than the Mobius transformations preserve symmetry? Linear fractional transformations preserve symmetric points, e.g., if the real axis in the z-plane gets mapped to the imaginary axis in the w-plane, then points symmetric with respect to the real axis get mapped to points that are again symmetric - with respect to the imaginary axis, which is the image of the real line.
Does such a useful symmetry argument exist for non-LFTs?
Thanks,
 A: If you have any meromorphic function $f:\mathbb{C}\to\mathbb{C}\cup\{\infty\}$ (not necessarily LFT) that maps the real axis to the imaginary axis, then consider $g(x) = f(x)/i$, which maps the real axis onto itself.  Then consider $h(x)=\overline{g(\overline{x})}$, which is another meromorphic function that agrees with $g$ on the whole real axis.  It follows that $h$ must equal $g$ everywhere.  Why: Identity Theorem (refer to "An Improvement"). [Edit: The theorem as stated in the link applies to holomorphic functions, not meromorphic functions.  To get it for meromorphic functions: given any desired point $x$, you can find some connected open set $U$ containing $x$ and intersecting the real axis and avoiding all the poles of $g$ and $h$, and then you can apply the holomorphic version of the theorem on $U$ to deduce that $g(x) = h(x)$.  I had originally included this explanation in my answer (honest!), but I inadvertently deleted it.]
Now we know $f(x)/i = \overline{f(\overline{x})/i}$, i.e., $f(\overline{x}) = -\overline{f(x)}$, from which your symmetry property follows.
TL;DR: Any meromorphic function on $\mathbb{C}$ that maps the real axis to the imaginary axis has the same symmetry property.
