conformal-field-theorycovariancediffeomorphism-invariancedifferential-geometrysymmetry
The title sums it pretty much. Are all diffeomorphism transformations also conformal transformations?
If the answer is that they are not, what are called the set of diffeomorphisms that are not conformal?
General Relativity is invariant under diffeomorphisms, but it certainly is not invariant under conformal transformations, if conformal transformations where a subgroup of diff, you would have a contradiction. Or I am overlooking something important?
It's somewhat unclear how you "understood" and are "happy" about the definition of the conformal transformations because your questions, while referring to page 2 and other things, are nothing else than misunderstandings about the definition of a conformal transformation which is explained on page 1, not 2.
John wrote his equation 1 which states that conformal transformations "are" diffeomorphisms that only change the metric by a local scalar coefficient - by a Weyl rescaling. However, the invariance of a theory under these diffeomorphisms is a trivial property. If one is allowed to change the terms coupled to the "metric", a diffeomorphism-transformed theory has a different action in general, and it is always possible to rewrite the original action in the conformally transformed coordinates.
But what's physically nontrivial is the condition that the action, in its original form, is actually invariant under the operations - that's what we mean by the theory's being conformally invariant. If a theory is conformally invariant, we don't allow any "change of the coefficients" in the integral of the Lagrangian density. This condition of conformal invariance, as he shows, is equivalent to the invariance of the theory under the "Weyl rescaling" only: we just completely eliminate the diffeomorphisms from the picture.
So:
Again, the invariance under the combined "diffeomorphism" and "Weyl rescaling" (the latter changes the form of the action) is a tautology. Obviously, by conformal invariance, we don't mean a tautology, so by conformal invariance, we mean the invariance of the action under the diffeomorphism separately, without changing the form of the metric in the action. Because the invariance under the "combo" is tautological, conformal invariance is equivalent to the invariance under the "Weyl rescaling part" of the transformation only.
No, on page 2, the transformations are exactly what the equations say: point-dependent transformations of the metric tensor itself i.e. a Weyl rescaling. There is no diffeomorphism at this stage. The invariance under those Weyl transformations is equivalent to the invariance under some diffeomorphisms (with the modification of the metric erased), as explained in the previous point.
The formalism may depend on classical physics but your comment that it is "classically only of course" is incorrect, too. All those facts about conformal transformations are completely valid quantum mechanically as well - and indeed, this background is primarily mean to understand some quantum theories.
Let there be given a 4-dimensional real manifold$^1$ $M$. As OP says, the set ${\rm Diff}(M)$ is the group of globally defined $C^{\infty}$-diffeomorphisms $f:M\to M$. The set ${\rm Diff}(M)$ is an infinite-dimensional Lie group. (To actually explain mathematically what the previous sentence means, one would have to define what an infinite-dimensional manifold is, which is beyond the scope of this answer.)
There is also the groupoid ${\rm LocDiff}(M) \supseteq {\rm Diff}(M)$ of locally defined $C^{\infty}$-diffeomorphisms $f:U\to V$ (i.e. the invertible morphisms in the category). Here $U,V\subseteq M$ are open neighborhoods (i.e. objects in the category).
The above is part of the active picture. Conversely, in the passive picture, there is the groupoid $LCT(M)$ of local coordinate transformations $f:U\to V$, where $U,V\subseteq\mathbb{R}^4$. Heuristically, due to the dual active & passive formulations, the two groupoids ${\rm LocDiff}(M)$ and $LCT(M)$ must be closely related at "the microscopic level". (We leave it to the reader to try to make the previous sentence precise.)
Next let us consider the frame bundle $F(TM)$ of the tangent bundle $TM$. It is a principal bundle with structure group $GL(4,\mathbb{R})$, which is a 16-dimensional Lie group.
Given two locally defined sections $$(e_0, e_1, e_2, e_3), (e^{\prime}_0, e^{\prime}_1, e^{\prime}_2, e^{\prime}_3)~\in~ \Gamma(F(TM_{|W})), \tag{1}$$ in some neighborhood $W\subseteq M$, then there is a locally defined $GL(4,\mathbb{R})$-valued section $$\Lambda~\in~\Gamma(GL(4,\mathbb{R})\to W), \tag{2}$$ such than the two sections (1) are related via$^2$ $$e^{\prime}_{b}~=~ \sum_{a=0}^3 e_{a} \Lambda^{a}{}_{b}, \qquad b~\in~\{0,1,2,3\} \tag{3}. $$
Conversely, given only one of the sections in eq. (1), we can use an arbitrary $GL(4,\mathbb{R})$-valued section (2) to define the other frame via eq. (3).
Let us now return to the groupoid $LCT(M)$ and see how the structure group $GL(4,\mathbb{R})$ comes in. In details, let there be given two local coordinate charts $U,U^{\prime}\subseteq M$, with non-empty overlap $U\cap U^{\prime}\neq \emptyset$, and with local coordinates $(x^0,x^1,x^2,x^3)$ and $(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})$, respectively. Then we have two locally defined sections $$\left(\frac{\partial}{\partial x^0}, \frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2},\frac{\partial}{\partial x^3}\right)~\in~ \Gamma(F(TM_{|U})), $$ $$\left(\frac{\partial}{\partial x^{\prime 0}}, \frac{\partial}{\partial x^{\prime 0}}, \frac{\partial}{\partial x^{\prime 0}},\frac{\partial}{\partial x^{\prime 0}}\right)~\in~ \Gamma(F(TM_{|U^{\prime}})), \tag{4}$$ in the frame bundle. The analogue of the $GL(4,\mathbb{R})$-valued section (2) is given by the (inverse) Jacobian matrix$^1$ $$ \Lambda^{\mu}{}_{\nu} ~=~\frac{\partial x^{\mu}}{\partial x^{\prime \nu}},\qquad \mu,\nu~\in~\{0,1,2,3\}, \tag{5} $$ cf. the chain rule.
Conversely, note that not all $GL(4,\mathbb{R})$-valued sections (2) are of the form of a Jacobian matrix (5). Given one local coordinate system $(x^0,x^1,x^2,x^3)$ and given a $GL(4,\mathbb{R})$-valued section (2), these two inputs do not necessarily define another local coordinate system $(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})$. The $GL(4,\mathbb{R})$-valued section (2) in that case evidently needs to satisfy the following integrability condition $$ \frac{\partial (\Lambda^{-1})^{\nu}{}_{\mu}}{\partial x^{\lambda}} ~=~ (\mu \leftrightarrow \lambda). \tag{6}$$
So far we have just discussed an arbitrary $4$-manifold $M$ without any structure. For the rest of this answer let us consider GR, namely we should equip $M$ with a metric $g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$ of signature $(3,1)$.
Similarly, we introduce a Minkowski metric $\eta_{ab}$ in the standard copy $\mathbb{R}^4$ used in the bundle $GL(4,\mathbb{R})\to W$. We now restrict to orthonormal frames (aka. as (inverse) tetrads/vierbeins) $$(e_0, e_1, e_2, e_3)~\in~ \Gamma(F(TM_{|W})), \tag{7}$$ i.e. they should satisfy the orthonormal condition $$ e_a\cdot e_b~=~\eta_{ab}. \tag{8} $$
Correspondingly, the structure group $GL(4,\mathbb{R})$ of the frame bundle $F(TM)$ is replaced by the proper Lorentz group $SO(3,1;\mathbb{R})$, which is a $6$-dimensional Lie group. In particular the sections (2) are replaced by locally defined $SO(3,1;\mathbb{R})$-valued sections $$\Lambda~\in~\Gamma(SO(3,1;\mathbb{R})\to W). \tag{9}$$ This restriction is needed in order to ensure the existence of finite-dimensional spinorial representations, which in turn is needed in order to describe fermionic matter in curved space. See also e.g. this Phys.SE post and this MO.SE post.
Consider a covariant/geometric action functional $$S[g, \ldots; V]~=~\int_V \! d^4~{\cal L} \tag{10}$$ over a spacetime region $V\subseteq M$, i.e. $S[g, \ldots; V]$ is independent of local coordinates, i.e. invariant under the groupoid $LCT(M)$. The action $$S[g, \ldots; V]~=~S[f^{\ast}g, \ldots; f^{-1}(V)] \tag{11}$$ is then also invariant under pullback with locally defined diffeomorphisms $f\in{\rm LocDiff}(M)$.
In summary, the symmetries of GR are:
--
$^1$ In most of this answer, we shall use the language of a differential geometer where e.g. a point/spacetime-event $p\in M$ or, say, a worldline have an absolute geometric meaning. However, the reader should keep in mind that a relativist would say that two physical situations which differ by an active global diffeomorphism are physically equivalent/indistinguishable, and hence a point/spacetime-event $p\in M$ does only have an relative geometric meaning.
$^2$ Conventions: Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while Roman indices $a,b,c, \ldots,$ are so-called flat indices.
Best Answer
A general diffeomorphism is not part of the conformal group. Rather, the conformal group is a subgroup of the diffeomorphism group. For a diffeomorphism to be conformal, the metric must change as,
$$g_{\mu\nu}\to \Omega^2(x)g_{\mu\nu}$$
and only then may it be deemed a conformal transformation. In addition, all conformal groups are Lie groups, i.e. with elements arbitrarily close to the identity, by applying infinitesimal transformations.
Example: Conformal Group of Riemann Sphere
The conformal group of the Riemann sphere, also known as the complex projective space, $\mathbb{C}P^1$, is called the Möbius group. A general transformation is written as,
$$f(z)= \frac{az+b}{cz+d}$$
for $a,b,c,d \in \mathbb{C}$ satisfying $ad-bc\neq 0$.
Example: Flat $\mathbb{R}^{p,q}$ Space
For flat Euclidean space, the metric is given by
$$ds^2 = dz d\bar{z}$$
where we treat $z,\bar{z}$ as independent variables, but the condition $\bar{z}=z^{\star}$ signifies we are really on the real slice of the complex plane. A conformal transformation takes the form,
$$z\to f(z)\quad \bar{z}\to\bar{f}(\bar{z})$$
which is simply a coordinate transformation, and the metric changes by,
$$dzd\bar{z}\to\left( \frac{df}{dz}\right)^{\star}\left( \frac{df}{dz}\right)dzd\bar{z}$$
as required to ensure it is conformal. We can specify an infinite number of $f(z)$, and hence an infinite number of conformal transformations. However, for general $\mathbb{R}^{p,q}$, this is not the case, and the conformal group is $SO(p+1,q+1)$, for $p+q > 2$.