$\DeclareMathOperator\SO{SO}\DeclareMathOperator\SL{SL}$It is clear that $\SO(3,\mathbb{R}) \subset \SL(3,\mathbb{R})$ acts isometrically, and thus by Moebius transformations, on the sphere at infinity.
Moreover, any $g \in SL(3,\mathbb{R})$ can be written, by the polar decomposition, as
$$g = pq$$
where $q \in SO(3)$ and $p$ is a real symmetric positive-definite $3$-by-$3$ matrix of determinant $1$. So the action of $g$ on the conformal structure $c$ is the same as the action of $p$ on the conformal structure $c$.
I claim that if $p \neq \mathbf{1}$, then $p^*(c) \neq c$, i.e. $p$ (and therefore $g$) does not induce a Moebius transformation on the sphere at infinity.
To see that, note that one can always find an orthonormal basis of $\mathbb{R}^3$, with respect to which $p$ becomes diagonal, say $p$ becomes $\operatorname{diag}(\lambda_1, \lambda_2, \lambda_3)$. If $p \neq \mathbf{1}$, then there exist two diagonal elements which are not equal, say $\lambda_1 \neq \lambda_2$, WLOG. Consider then the point $N = (0,0,1)^T \in S^2_\infty$ (in the new coordinates). Then the Euclidean metric $g(N) = dx^2 + dy^2$ at the point $N$ gets pulled back by $p$ to $\lambda_1^2 dx^2 + \lambda_2^2 dy^2$ which is not a positive multiple of $g(N)$. This shows that the action of $p \neq \mathbf{1}$ is not conformal.
We thus conclude that the subgroup of $SL(3,\mathbb{R})$ which acts by Moebius transformations on the sphere at infinity is actually $SO(3,\mathbb{R})$, which actually acts isometrically on the $S^2_\infty$ with the round metric.
The orbit of $c$ is thus simply $SL(3,\mathbb{R})/SO(3,\mathbb{R})$.