Imposing that an action should be conformally invariant has a subtle but important difference with respect to imposing that it should be diffeomorphism invariant. Let me try to emphasize the differences between *Diffeomorphisms*, *Conformal transformations* and *Weyl transformations*. I will also clarify how to impose the invariance of the action, namely I'll argue whether the fields or the coordinates must be changed.

**Diffeomorphisms**

The action on the coordinates is given by a general differentiable function with differentiable inverse: $x^\mu \to {x'}^\mu = {x'}^\mu(x)$. The action on fields with spin 0, 1 and 2 is the following
$$
\begin{align}
\phi(x) &\to \phi'(x') = \phi(x)\,,
\\
\partial_\mu\phi(x) &\to \partial'_\mu\phi'(x') = \frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu\phi(x)\,,
\\
g_{\mu\nu}(x) &\to g'_{\mu\nu}(x') = \frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu}g_{\alpha\beta}(x)\,.
\end{align}
$$
These transformations are just changes of variables, every theory is invariant under them.

**Weyl transformations**

Weyl transformations do not act on the coordinates but simply rescale the fields
$$
\begin{align}
\phi(x) &\to \tilde{\phi}(x) = \Omega^{-\Delta}(x)\phi(x)\,,
\\
\partial_\mu\phi(x) &\to \partial_\mu\tilde{\phi}(x) = \partial_\mu(\Omega^{-\Delta}(x)\phi(x))\,,
\\
g_{\mu\nu}(x) &\to \tilde{g}_{\mu\nu}(x) = \Omega^{2}(x)g_{\mu\nu}(x)\,.
\end{align}
$$

**Conformal transformations**

Conformal transformations are coordinate transformations that preserve the metric up to a scale factor, that is $x^\mu \to {x'}^\mu$ and
$$
\frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu} g_{\alpha\beta}(x)= \Omega^{-2}(x)g_{\mu\nu}(x)\,.\tag{1}
$$
In OP's example ${x'}^\mu = \lambda x^\mu$ and $\Omega(x) = \lambda$. Now, this is the key point: we *don't* want to impose that the theory is invariant under this change of variables, this is always true because it is a particular case of diffeomorphism invariance. We want instead to compare theories with the same metric. That is, we always imagine a conformal transformation as a diffeomorphism satisfying (1) followed by a Weyl rescaling that cancels the $\Omega$ factor. Therefore the transformation laws are
$$
\begin{align}
\phi(x) &\to \tilde{\phi}'(x') = \Omega^{-\Delta}(x)\phi(x)\,,
\\
\partial_\mu\phi(x) &\to \partial'_\mu\tilde{\phi}'(x') = \frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu(\Omega^{-\Delta}(x)\phi(x))\,,
\\
g_{\mu\nu}(x) &\to \tilde{g}'_{\mu\nu}(x') = g_{\mu\nu}(x)\,.
\end{align}
$$
With the notation "tilde prime" I denote the composition of the two last transformations.

**Invariance of the action**

Consider a generic Lagrangian that depends on the field $\phi$, its first derivative and the metric tensor. The action then reads
$$
S = \int d^4x \, \mathcal{L}[g_{\mu\nu}(x),\phi(x),\partial_\mu\phi(x)]\,.
$$
The transformation acts *only* on the fields and leaves $x$ unchanged since it is a dummy variable (integrated over).
Proving diffeomorphism invariance amounts to requiring
$$
\begin{align}
S &= \int d^4x\, \mathcal{L}\left[g'_{\mu\nu}(x),\phi'(x),\partial'_\mu\phi'(x)\right] =
\\
&=\int d^4x'\, \mathcal{L}\left[g'_{\mu\nu}(x'),\phi'(x'),\partial'_\mu\phi'(x')\right] =
\\
&=\int d^4x\,|\Omega(x)|^4 \mathcal{L}\left[\Omega^{-2}(x)g_{\mu\nu}(x),\phi(x),\frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu\phi(x)\right]\,,
\end{align}
$$
in the second step I just renamed $x$ to $x'$, it's a trivial operation. In the last step I used the transformation properties of the measure and of the fields (See Di Francesco - Mathieu - Sénéchal for example).

A Weyl rescaling requires simply
$$
S = \int d^4x\, \mathcal{L}\left[\Omega^2(x)g_{\mu\nu}(x),\Omega^{-\Delta}(x)\phi(x),\partial_\mu(\Omega^{-\Delta}(x)\phi(x))\right]\,.
$$
Finally, the requirement for conformal invariance can be obtained by composing the two transformations above
$$
S = \int d^4x\, |\Omega(x)|^4\mathcal{L}\left[g_{\mu\nu}(x),\Omega^{-\Delta}(x)\phi(x),\frac{\partial x^\nu}{\partial{x'}^\mu}\partial_\nu(\Omega^{-\Delta}(x)\phi(x))\right]\,.
$$
This is consistent with what OP says. Notice that since diffeomorphism invariance is always true, just proving Weyl invariance is enough. The converse is not obvious and it is object of current research. We could say that by imposing conformal invariance we are imposing Weyl invariance only by those $\Omega$'s that make $g_{\mu\nu}$ and $\Omega^2 g_{\mu\nu}$ diffeomorphic.

## Best Answer

OK I think I know what is going on. It's all about primes. Consider an active spacetime transformation:

(the transformation of the metric tensor follows from the fact that it is a rank 2 tensor). With this notation both Di Francesco and David Tong are wrong (as far as I understand). The GR book by Zee on the other hand writes it properly. First of all consider an

isometry. This is an spacetime transformation as before that leaves the metric invariant, meaning(watch the primes). On the other hand a

conformal transformationis a transformation that satisfies a weaker condition: it leaves the metric invariant up to scale, meaningNow there should be no inconsistency. Di Francesco's definition was wrong (according to this convention/notation/understanding) because it compared the metric before and after the transformation at different points, and you have to compare them at the same point.