= Solution
By transitivity, choose $h\in G$ with $hy=x$. If $gy=y$, then
$$
(hgh^{-1})x=hgy=hy=x.
$$
Thus $hgh^{-1}\in B$. This proves the conjugation assertion directly from the <transitive group action>.
For the <complex special linear group in dimension two>, the <Möbius transformation> of a matrix $g$ acts by $z\mapsto(az+b)/(cz+d)$, with $ad-bc=1$. It fixes infinity exactly when $c=0$, so its <stabilizer subgroup> is
$$
\boxed{B=\left\{\begin{pmatrix}a&b\\0&a^{-1}\end{pmatrix}:a\in\mathbb C^\times,\ b\in\mathbb C\right\}.}
$$
We distinguish the cases to avoid dividing by a vanishing coefficient.
If $c\ne0$, infinity is not fixed, and the finite fixed points solve
$$
cz^2+(d-a)z-b=0.
$$
Neither root is the pole $-d/c$: substituting that value gives $(ad-bc)/c=1/c\ne0$. The fixed-point set is therefore
$$
\boxed{\left\{\frac{a-d+\sqrt{(a+d)^2-4}}{2c},\ \frac{a-d-\sqrt{(a+d)^2-4}}{2c}\right\},}
$$
with the two entries coinciding when the discriminant is zero. The discriminant identity uses $(a-d)^2+4bc=(a+d)^2-4$.
If $c=0$ and $a\ne d$, the fixed points are $\{\infty,b/(d-a)\}$. If $c=0$, $a=d$ and $b\ne0$, only infinity is fixed. If $c=0$, $a=d$ and $b=0$, the determinant condition gives $g=\pm I$ and every point is fixed. Both central matrices induce the identity Möbius transformation.
Every case has a fixed point, because a quadratic over the complex numbers has a root. The action is transitive: any finite $z_0$ is sent to infinity by
$$
h=\begin{pmatrix}0&-1\\1&-z_0\end{pmatrix}\in SL_2(\mathbb C),
$$
and infinity already needs no change. Applying the first argument to a fixed point proves \b[every element of $SL_2(\mathbb C)$ is conjugate to an element of $B$].
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