Let $X$ be a subset of $\mathbb{C}$ where the real axis is removed
$$X = \{z\in\mathbb{C}\mid\operatorname{Im}(z)\neq0\}.$$
Let $\mathcal{M}$ be the group of Mobius Transformations that send the real axis to itself.
Is $(X, \mathcal{M})$ homogenous?
As homogenous means that any two points in $X$ are congruent, this means that any two complex points need to be mapped to each other, which isn't possible, as we can't map any $z$ to $0$ and use the group structure of $\mathcal{M}$ to argue that all points are congruent, is this the correct train of thought?
$SL(2,\mathbb{R})$ acts on $\mathbb{C}$ (in fact it acts on the sphere,$S^{2}$) by Mobius Transformations $$A\in SL(2,\mathbb{R}) ,A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ $$Az=\frac{az+b}{cz+d}$$this action has three orbits,as u mentioned the real axis is already one orbit,the other two orbits being the upper half plane $Im(z)>0$ and the lower half plane .Reamrk that $SL(2,\mathbb{R})$ is connected and the action is continous ,So any orbit is coonected, in particular $SL(2,\mathbb{R})$ cannot act transitively on $X$, of course u can show ths in a more easier way by showing that the upper half plane is invariant , meaning that if $$\forall A \in SL(2,\mathbb{R}),Im(z)>0 \implies Im(Az)>0$$ (In particular the geometry u are asking for isn't homogeneous),of course u can show that the orbit of any real number is $\mathbb{R}$. $$ $$ Edit:Proof of the invariance of the upper half-plane by the action of $SL(2,\mathbb{R})$: let $z\in \mathbb{C}$ be such that $Im(z)>0$,let's compute the imaginary part of $Az$: $$Im(Az)=Im(\frac{az+b}{cz+d})=Im(\frac{(az+b)(c\overline{z}+d)}{\lvert cz+d\rvert})=\frac{Im(az+b)(c\overline{z}+d)}{\lvert cz+d\rvert}=\frac{Im(acz\overline{z}+bd+adz+bc\overline{z})}{\lvert cz+d\rvert}=\frac{Im(z)}{\lvert cz+d\rvert}>0$$ So if $Z$ belongs to the upper half plane ,so does its image.