thank you for the nice question. It directly relates to the topics of conformal field theories. I found a very nice thread in another forum where I guess your question has been answered.
Nevertheless, I will try to summarize the main points here and maybe add some points.
Symmetries in General Relativity
In general relativities, symmetries correspond to an isometry of the metric $g=g_{ab}dx^a dx^b$, say $\varphi^\star g = g$. That means, if you move along the path of such a symmetry, it does not change. This can be expressed in terms of the Lie-derivative.
$$L_v g = 0$$
or
$$\nabla_{(a}v_{b)} = 0$$
where the parenthesis stands for symmetrization over indices and $v = \dot\varphi(t)$ is the vector field associated to $\varphi$. One can find very nice introductory calculations for this in Robert M. Wald: General Relativity and Hans Stephani's Introduction to Special and General Relativity.
If $n$ is a unit geodesic, further integration of
$$Q=v_a n^a$$
leads to conserved quantities since
$$n^a \nabla_a \left(Q = n^b v_b\right) = n^a n^b\nabla_b v_a + v_b n^a \nabla_a n^b\equiv 0 $$
due to the Killing and geodesic equations.
Famous examples are mass $M$ (or energy) for a stationary spacetime or angular momentum $J$ for axial symmetry (yes, one can assign a spacetime an angular momentum, I found it puzzling in the first place),
$$M = 2\int_\Sigma \left( T_{ab}-\frac12 T^n_{\,n}g_{ab} \right)n^a \xi^b dV$$
$$J = -\int_\Sigma T_{ab}n^a \eta^b dV$$
where now $\xi$ is the stationary Killing vector, often $\xi = \partial_t$ and $\eta$, often $\eta=\partial_\varphi$ holds for the axial symmetry and $n$ is now vector perpendicular to a space-like hypersurface $\Sigma$.
Conformal isometries
Now, the situation is a little bit different. A conformal Killing vector $c$ now gives rise to a symmetry of the form
$$L_cg=\omega^2g$$
and the conformal Killing equation, implicitly defining $\omega$ now takes the form
$$\nabla_{(a}c_{b)} = \frac1n g_{ab}\nabla_d c^d$$
In your case, you force $\omega = 1$ but this is not of great importance as you will see next.
What happens to the "conservation equation"? We have
$$n^a \nabla_a \left( n^b c_b\right) = \frac1n \left( \nabla_d c^d \right) n^a n_a$$
which is only zero if $n^a n_a = 0$, a null-geodesic. So, only for a very special class of movements, here light-particles, one will find a symmetry. But this was expected since conformal transformations will not change angles thus light movement won't be affected.
I don't think that this is a conserved quantity in the sense of Emmy Noether.
Sincerely
Robert
PS.: I apologize for any inconvenience concerning notation. I hope everything is clear from context.