Simplifying effect of a hidden Weyl symmetry in a QFT on curved spacetime We consider AdS$_{d+1}$ in Poincaré coordinates:
$$
ds^2=\frac{1}{z^2}\left(-dt^2+dz^2+dx_{d-1}^2\right),
$$
where we set the AdS radius to unity. We study a scalar in this background with action
$$
S=-\frac{1}{2}\int d^{d+1}x\,\sqrt{-g}\left(\partial_\mu\phi\partial^\mu\phi+m^2\phi^2\right).
$$
I want to understand the following statement:

For $m^2=-\frac{(d-1)(d+1)}{4}$, there is a  hidden Weyl symmetry present in the action which simplifies the form of the scalar correlators in the vacuum state.

What I was able to show is the following: If we consider a scalar with a different action
$$
S=-\frac{1}{2}\int d^{d+1}x\,\sqrt{-g}\left(\partial_\mu \phi \partial^\mu \phi+ \frac{d-1}{4d}R\phi^2\right).
$$
i.e. with coupling to the scalar curvature instead of a mass term, then this action is indeed invariant under Weyl transformations
$$
\begin{align}
g_{\mu\nu} &\to \Omega^2 g_{\mu\nu}, \\
 \phi &\to \Omega^{\frac{1-d}{2}}\phi,
\end{align}
$$
with a spacetime dependent $\Omega=\Omega(x)$.
For AdS$_{d+1}$ the Ricci scalar is $R=-d(d+1)$, so with a non-dynamical AdS background one can rewrite the action as
$$
S=-\frac{1}{2}\int d^{d+1}x\,\sqrt{-g}\left(\partial_\mu \phi \partial^\mu \phi- \frac{(d-1)(d+1)}{4}\phi^2\right),
$$
so one finds the massive scalar with the $m^2=-\frac{(d-1)(d+1)}{4}$.


*

*Now, when gravity is not dynamical and the Ricci scalar is not explicit in the action any more, how does this Weyl symmetry of the gravitational theory affect the scalar theory? 

*Does the hidden Weyl symmetry lead to a conformal symmetry of the scalar theory?
 A: Upshot:
The following two facts can be used to argue that the scalar correlator simplifies in the special case described above.


*

*When the scalar action is Weyl invariant, then the scalar equation of motion is covariant and we can use a Weyl transformation to simplify the equation.

*There is a Weyl transformation that maps AdS to the upper half plane of flat Minkowski space. The correlator in flat space can be computed easily and the result can be transformed back to AdS using the inverse of the Weyl transformation above.


More details:
In the special case of $m^2=-\frac{(d-1)(d+1)}{4}$ the scalar action in Weyl invariant, i.e. under
$$
\begin{align}
g_{\mu\nu} &\to \Omega^2 g_{\mu\nu}, \\
 \phi &\to \Omega^{\frac{1-d}{2}}\phi,
\end{align}
$$
and the equation of motion
$$
\left(\Box-\frac{d-1}{4d}R\right)\phi=0
$$
is covariant.
For $\Omega=z$, the transformed metric is $z^2g_{\mu\nu}=\eta_{\mu\nu}$, the flat $(d+1)$ dimensional Minkowski metric. Defining the rescaled field $\varphi=z^{\frac{1-d}{2}}\phi$, the equation of motion becomes
$$
0=\eta^{\mu\nu}\partial_\mu\partial_\nu\varphi=\left(-\partial_t^2+\vec{\nabla}^2+\partial_z^2\right)\varphi.
$$
The solution is
$$
u=Ae^{-i\omega t+i\vec{k}\cdot\vec{x}+iqz}+Be^{-i\omega t+i\vec{k}\cdot\vec{x}-iqz},
$$
where $q=\sqrt{\omega^2-\vec{k}^2}$. The modes with $\vec{k}^2>\omega^2$ are forbidden as they are not normalizable in the region $z\to\infty$. Imposing Dirichlet boundary conditions $\left.\varphi\right|_{z=0}=0$ leads to
$$
u=C e^{i{k}\cdot{x}}\sin(qz)
$$
where $k\cdot x=-\omega t+\vec{k}\cdot\vec{x}$. Now, one can expand the field $\varphi$ in the eigenmodes $u$ we found. We separate positive and negative frequency modes:
$$
\varphi(x,z)=\int_{k^0>|\vec{k}|} d^d k\, \left(C a_{k} e^{i{k}\cdot{x}}\sin(qz)+C^* a_k^\dagger e^{-i{k}\cdot{x}}\sin(qz)\right).
$$
From this point on, quantizing and computing the two-point function is straightforward. First, one computes the Klein-Gordon norm of the mode functions to find the normalization $C$. Canonical commutation relations for $a_k$ and $a_k^\dagger$ are imposed and the vacuum two-point function can be found.
The result can be transformed back to $AdS$ using
$$
\langle\phi(x,z)\phi(y,z')\rangle= z^{-\frac{1-d}{2}}{z'}^{-\frac{1-d}{2}}\langle\varphi(x,z)\varphi(y,z')\rangle
$$
The result,
$$
\langle\varphi(x,z)\varphi(x',z')\rangle=\frac{\Gamma\left(\frac{d-1}{2}\right)}{4\pi^{\frac{d+1}{2}}}\left(\frac{1}{[(x-x')^2+(z-z')^2]^{\frac{d-1}{2}}}-\frac{1}{[(x-x')^2+(z+z')^2]^{\frac{d-1}{2}}}\right)
$$
is very interesting. Translational invariance along $x$ dictates that the correlator can only depend on $x-x'$. Because of the Dirichlet boundary conditions at $z=0$, which is where the AdS boundary is mapped under the rescaling, translation invariance in the $z$-direction is broken. In addition to a term depending on $z-z'$, there is a term $z+z'$. This can be compared to electromagnetism, where the Dirichlet boundary conditions can be enforced by introducing mirror charges. $-z'$ is basically a mirror image of $z'$ upon reflection at $z=0$, so there is a second term due to the Dirichlet boundary conditions on the AdS boundary.
A: A Weyl transformation is a conformal transformation. They just call it a Weyl transformation because its within General Relativity. This mean that your attempt to understand the statement (i.e. questions "1." and "2.") is invalid and has little to do with the relationship between the Weyl symmetry and conformal scalar symmetry. The statement you are attempting to understand says that the Weyl symmetry simplifies the scalar correlators i.e. when you are taking the vacuum expectation value of a massless scalar field.
e.g.
\begin{equation} \langle\phi(x_1),\phi(x_2)...\phi(x_n)\rangle= \frac{\int D\phi e^{-S[\phi]} \phi(x_1)...\phi(x_n)}{\int D\phi e^{-S[\phi]}} \end{equation} 
As you can see here taking the VEV is far simpler because
\begin{equation} S[\phi]=-\frac{1}2 \int d^{d+1}x \sqrt{-g}(\partial_u\phi\partial^u\phi-\frac{(d-1)(d+1)}4\phi^2) \end{equation}
