QFT in curvilinear coordinates I have a question that I believe is confusing me more than it should. We all know the path integral in the usual $(t,\vec{x})$ coordinates. For example, consider a simple $U(1)$ gauge theory. The propagator is given in the path integral formalism as
$$
\left< A_\mu(x_1) A_\nu(x_2) \right> = \int [d A_\alpha] \exp\left[ -\frac{1}{4} \int d^4 x F_{\mu\nu} F^{\mu\nu} \right] A_\mu(x_1) A_\nu(x_2)
$$
Of course, we still have to gauge fix, but let us not worry about this yet. The generalization to curved spacetime is
$$
\left< A_\mu(x_1) A_\nu(x_2) \right> = \int [d A_\alpha] \exp\left[ -\frac{1}{4} \int d^4 x \sqrt{-g} F_{\mu\nu} F^{\mu\nu} \right] A_\mu(x_1) A_\nu(x_2)
$$
Now suppose, instead of being in curved coordinates, we are simply in Minkowski space but using some funny coordinates. My question is, how does the 2-point function in curvilinear coordinates relate to that in the usual coordinates? Is it just
$$
\left< A'_\mu(x'_1) A'_\nu(x'_2) \right> = \frac{\partial x^\alpha }{\partial x'^\mu}\bigg|_{x=x_1} \frac{\partial x^\beta}{\partial x'^\nu}\bigg|_{x=x_2}\left< A_\mu(x_1) A_\nu(x_2) \right> 
$$
 A: Note. Substantial rewriting performed on 2014-05-31.
I believe the answer is yes without the qualification of diffeomorphism-invariance as I had originally claimed.
The proof.
Consider a QFT described by a classical action $S$ and containing a 4-vector field $A$.  For any diffeomorphism $f$ on $\mathbb R^{3,1}$, we define a transformed field $A_f$ in the standard way;
\begin{align}
  A_f^\mu(f(x)) = \frac{\partial f^\mu}{\partial x^\alpha}(x) A^\alpha(x).
\end{align}
Moreover, for any functional $\mathscr F$ of this vector field, we define it's expectation value as the following functional integral:
\begin{align}
  \langle \mathscr F\rangle = \int [dA] e^{-S[A]} \mathscr F[A].
\end{align}
The correlation functions $\langle A^\mu(x_1)A^\nu(x_2)\rangle$ and $\langle A_f^\mu(f(x_1))A_f^\nu(f(x_2))\rangle$ are defined as special cases of this construction by computing the expectation values of the functionals
\begin{align}
  \mathscr C^{\mu\nu}(x_1, x_2)[A] &:= A^\mu(x_1)A^\nu(x_2),\\
  \mathscr C_f^{\mu\nu}(x_1, x_2)[A] &:= A_f^\mu(f(x_1))A_f^\nu(f(x_2)),
\end{align}
so we have
\begin{align}
  \langle A_f^\mu(f(x_1))A_f^\nu(f(x_2))\rangle 
&= \int [dA] e^{-S[A]} A_f^\mu(f(x_1))A_f^\nu(f(x_2)) \\
&= \int [dA] e^{-S[A]} \frac{\partial f^\mu}{\partial x^\alpha}(x_1) A^\alpha(x_1)  \frac{\partial f^\nu}{\partial x^\beta}(x_2) A^\beta(x_2)\\
&= \frac{\partial f^\mu}{\partial x^\alpha}(x_1)\frac{\partial f^\nu}{\partial x^\beta}(x_2)\int [dA] e^{-S[A]}  A^\alpha(x_1)   A^\beta(x_2)\\
&= \frac{\partial f^\mu}{\partial x^\alpha}(x_1)\frac{\partial f^\nu}{\partial x^\beta}(x_2)\langle A^\mu(x_1)A^\nu(x_2)\rangle
\end{align}
In other words, the desired result follows simply from linearity of the functional integral.
A Note on diffeomorphism invariance.
In my original answer, I had claimed that diffeomorphism invariance was necessary for the above result, but I believe that was wrong, as the above computation shows.
However, I do have something relevant to say about diffeomorphism invariance.  What we proved above is that the two-point function transforms as a two-tensor.  That did not require diffeomorphism invariance; diffeomorphism invariance gives us a different constraint on the two-point function.  
Notice that if the functional integration measure $[dA]$ is diffeomorphism invariant, namely if $[dA_f] = [dA]$, and if the classical action is diffeomorphism invariant, namely if $S[A_f] = S[A]$, then in addition to the computation above, we also have
\begin{align}
  \langle A_f^\mu(f(x_1))A_f^\nu(f(x_2))\rangle
&= \int [dA] e^{-S[A]} A_f^\mu(f(x_1))A_f^\nu(f(x_2)) \\
&= \int [dA_f] e^{-S[A_f]} A_f^\mu(f(x_1))A_f^\nu(f(x_2)) \\
&= \int [dA] e^{-S[A]} A^\mu(f(x_1))A^\nu(f(x_2)) \\
&= \langle A^\mu(f(x_1))A^\nu(f(x_2))\rangle. \tag{$\star$}
\end{align}
If we combine this with the property proven above about the two-tensor nature of the two-point function, then we have the following constraint which encodes diffeomorphism invariance:
\begin{align}
  \frac{\partial f^\mu}{\partial x^\alpha}(x_1)\frac{\partial f^\nu}{\partial x^\beta}(x_2)\langle A^\mu(x_1)A^\nu(x_2)\rangle = \langle A^\mu(f(x_1))A^\nu(f(x_2))\rangle
\end{align}
Compare to eq. (2.148) of Di-Franceso et. al. Conformal Field Theory.
The main point is that two-tensor transformation of the two-point function is an immediate consequence of linearity of the functional integral, but if we want a property like $(\star)$, then we also need diffeomorphism invariance of the classical action and the functional integration measure.
