Preliminaries
Any symmetry transformation acts ONLY on the fields, never ever on the coordinates. Often, transformations of coordinates is used as a TOOL to describe how the fields themselves transform. For instance, under translations, we have
$$
U(a) \phi(x)U(a)^{-1} = \phi'(x).
$$
Notice that the transformation has not acted on the coordinates. However, the coordinate transformation is often invoked to describe $\phi'$ in terms of $\phi$ as
$$
\phi(x) = \phi'(x+a) .
$$
This does not mean that coordinates were acted upon by $U(a)$. The way we should truly interpret this formula is that
$$
\phi'(x) = \sum_{k=0}^\infty \frac{(-1)^k}{k!} a^{\mu_1} \cdots a^{\mu_k} \partial_{\mu_1} \cdots \partial_{\mu_n} \phi(x) .
$$
so that everything on both sides is at the same point $x$.
Every field theory has some dynamical fields $\phi(x)$ and some background fields ${\bar \phi}(x)$. Symmetry transformations act ONLY on dynamical fields. Background fields are to be treated as ${\mathbb C}$-numbers on the Hilbert space so for example
$$
U \phi(x) U^{-1} = \phi'(x) , \qquad U {\bar \phi}(x) U^{-1} = {\bar \phi}(x) .
$$
One of the many background fields are the metric $g_{\mu\nu}(x)$. In many cases (especially at a beginner level), the metric is the only background field. For simplicity, I will assume that this is the case in the rest of this post.
Diffeomorphisms
Under diffeomorphisms, the fields transform as
$$
\phi(x) \to \phi'(x) , \qquad g_{\mu\nu}(x) \to g'_{\mu\nu}(x)
$$
where
$$
\phi(x) = D \left( \frac{ \partial f^\mu(x) }{ \partial x^\nu} \right) \cdot \phi'(f(x)) , \qquad g_{\mu\nu}(x) = \frac{ \partial f^\alpha(x) }{ \partial x^\mu} \frac{ \partial f^\beta(x) }{ \partial x^\nu} g'_{\alpha\beta}(f(x))
$$
Here, $D$ is the representation of $GL(d,{\mathbb R})$ under which the field $\phi$ transforms.
The action is invariant under such a transformation. However, this is NOT a symmetry of the theory since the background fields are also transforming. However, if we consider only a subset of diffeomorphisms $f^\mu(x)$ such that
$$
g'_{\mu\nu}(x) = g_{\mu\nu}(x)
$$
then $f^\mu(x)$ generate symmetries of the theory. These are called isometry transformations
Weyl Transformations
Diffeomorphisms preserve the action in ANY local field theory. Some quantum field theories also admit another transformation which preserves the action known as Weyl transformations under which
$$
\phi(x) \to \phi_\Omega(x) , \qquad g_{\mu\nu} \to \Omega(x)^2 g_{\mu\nu}(x)
$$
It is often the case that $\phi_\Omega(x) = \Omega(x)^{-\Delta} \phi(x)$, but this may not always be true (e.g. stress tensor in two dimensions).
Weyl transformations are NEVER symmetries of the theory since the background field always transforms.
Conformal Transformations
While Weyl transformations are NOT symmetries of the theory, its existence as an invariant transformation of the action allows us to EXTEND the isometry symmetry we discussed earlier. We do this by considering the following sequence of transformations
$$
g_{\mu\nu}(x) \stackrel{\text{diff}}{\to} \Omega_f(x)^2 g_{\mu\nu}(x) \stackrel{\text{Weyl}}{\to} g_{\mu\nu}(x)
$$
In this way, this composite transformation IS a symmetry of the theory since the background field are invariant. This composite transformation is known as "conformal transformation".
Summary
In a summary, a conformal transformation involves TWO transformations - a diffeomorphism which transforms the metric by a conformal factor and a Weyl transformation that removes that factor.
In other words, to answer OPs question - a conformal transformation consists of both a (special) diffeomorphism AND a Weyl transformation. However, in many texts, the diffeomorphism part of a conformal transformation is -- unfortunately and mistakenly -- also referred to as a conformal transformation.
