Does Einstein field equation imply there is always conformal symmetry in 2 dimension? If we contract both sides of the Einstein field equation $$R_{\mu \nu }-{\frac {1}{2}}Rg_{\mu \nu }=-{8\pi G}T_{\mu \nu },$$ we will get $$\frac{2-n}{2}R=-8\pi G T_\mu^\mu,$$ where $n$ is the dimension of spacetime. Therefore in 2 dimensional spacetime we always have $T_\mu^\mu=0$, which implies any 2d field theory has conformal symmetry, at least classically. But this statement sounds unreasonable. What's wrong here?
 A: Analogy: consider an infinitesimal symmetry transformation. Noether's theorem states that the current is conserved if the action doesn't change. But according to the least action principle the infinitesimal variation of the action should always be zero in classical theory! Conclusion: any current is conserved.
This is obviously wrong. But why?
Spoiler alert: if you like puzzles, you might want to try to figure this out on your own. Below is the answer, don't read it unless you want to :)
The answer to this is, of course, that the current is conserved if the symmetry doesn't change the action off-shell. That is, on any classical configuration, not necessarily satisfying the equations of motion.
A moment of reflection would convince you that the same thing happens here. A field is only conformal-invariant is $T_{\mu}^{\mu}$ vanishes off-shell. That is, without the implication of equations of motion. Einstein's equations have nothing to do with this.
Try checking that stress-energy trace vanishes off-shell for a massless scalar in 2d.
P.S. And yes, I believe Slereah's comment is wrong and misleading.
