How does the 2D free boson transform under global conformal transformations? Consider the 2D free boson
$$\mathcal{L}=\int d^2 z\ \partial\phi\ \bar{\partial}\phi$$
How does $\hat{\phi}(z)$ transform under a global conformal transformation $\Lambda=\begin{pmatrix}a & b\\ c & d\end{pmatrix}$? I think that
$$U[\Lambda]\hat{\phi}(z)U^{\dagger}[\Lambda]=\hat{\phi}\left(\frac{az+b}{cz+d}\right)$$
However if this were the case then the two-point function
$$M(z,w)=\langle \hat{\phi}(z)\hat{\phi}(w)\rangle$$
would satisfy
$$M\left(\frac{az+b}{cz+d},\frac{aw+b}{cw+d}\right)=M(z,w)$$
which the well known two-point function
$$\langle \hat{\phi}(z)\hat{\phi}(w)\rangle=\ln|z-w|^2$$
clearly does not satisfy. Question: what is $U[\Lambda]\hat{\phi}(z)U^{\dagger}[\Lambda]$?
 A: I assume you've read that in the 2D free boson theory, you're "supposed" to work only with operators like $\partial \phi$ and $e^{i\alpha \phi}$, avoiding $\phi$ itself. But if you're trying to consider the more general theory anyway to see what happens, we need to understand why using
\begin{equation}
\phi^\prime(x^\prime) = \left | \frac{\partial x^\prime}{\partial x} \right |^{-\Delta/d} \phi(x), \;\;\;\; (**)
\end{equation}
as described in the last answer, leads to the contradiction $\left < \phi(z) \phi(w) \right > = 1 \neq \log(z - w)$. The reason is that (**) assumes $\phi$ to be an eigenstate of the dilation operator. But since all powers of $\phi$ have dimension zero, this does not come automatically... we need to check if it's true.
It is helpful to define $\mathcal{O}_n = \frac{1}{\sqrt{n!}} \phi^n$ and start with the Wick-like result
\begin{equation}
\left < \mathcal{O}_n(z) \mathcal{O}_n(w) \right > = \log^n(z - w)
\end{equation}
and hit it with the differential operator that appears in the Ward identity for the generator of dilations. This leads to
\begin{equation}
\left [ z \partial_z + w \partial_w \right ] \left < \mathcal{O}_n(z) \mathcal{O}_n(w) \right > = n \log^{n - 1}(z - w).
\end{equation}
Comparing to equation (2.5) of https://arxiv.org/abs/1605.03959 we see that this is compatible with dilations acting on the $\mathcal{O}_n$ according to
\begin{equation}
i[D, \mathcal{O}_n(z)] = \left ( \Delta_n^{\;\;m} + \delta_n^m z \partial_z \right ) \mathcal{O}_m(z)
\end{equation}
as long as the scaling matrix is given by
\begin{equation}
\Delta = \begin{pmatrix}
0 & 1 & 0 & 0 & \dots \\
0 & 0 & 2 & 0 & \dots \\
0 & 0 & 0 & 3 & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{pmatrix}.
\end{equation}
The pesky thing about this matrix isn't only that it's not diagonal but that it's not even diagonalizable. In fact, these powers of $\phi$ form an indecomposable representation with an infinitely large Jordan block. So their transformation properties all have to be considered together.
In two dimensions especially, these so called logarithmic CFTs have been widely studied but usually in cases with finite Jordan blocks. Included in this are models of percolation and systems with quenched disorder as described in https://arxiv.org/abs/1303.0847 and other reviews. It is probably possible to go further with analyzing the free boson CFT in this way as well. It's just not going to be found in introductory textbooks.
A: Let's define conformal transformation first in active point of view.
We have $$x\to x'$$ and
$$\phi(x)\to\phi'(x')$$ we also define a functional such that
$$\phi'(x')=\mathcal{F}(\phi(x))$$
So in order to understand better let's see several explicit examples,
Translation:
$$x'=x+a$$
$$\phi'(x')=\phi(x)$$ thus in this case $\mathcal{F}$ is trivial.
Lorentz:
$$x^{'\mu}=\Lambda^{\mu}_\nu x^\nu$$ and
$$\phi'(\Lambda x)=L_\Lambda\phi(x)$$ where $L$ is the representation of Lorentz group. Thus in this case $F$ is $L$.
Now let's return to your case, one can easily show that we have the following identity,
$$\langle\phi(x')\phi(y')\rangle=\langle\mathcal{F}(\phi(x))\mathcal{F}(\phi(y))\rangle$$
so thats how you should transform the correlation functions. The general conformal transformation  for spinless fields is given as,
$$\phi'(x')=|\frac{\partial x'}{\partial x}|^{-\Delta/d}\phi(x)$$
where $|\frac{\partial x'}{\partial x}|$ is jacobian. This holds if the action is invariant under those transformations.
