Shouldn't every co-ordinate transformation preserve angles between vectors? If yes, then why is conformal transformation special? Consider the dot product in a metric space between two vectors. Below I show that the dot product doesn't change by a co-ordinate transformation.
$g_{\mu \nu} A^{\mu} B^{\nu} = g_{\mu \nu} \frac{\partial x_{\mu}}{\partial x^{\prime}_{\alpha}} {A^\prime}^{\alpha} \frac{\partial x_{\nu}}{\partial x^{\prime}_{\beta}} {B^{\prime}}^{\beta}$
Since $g_{\mu \nu} \frac{\partial x_{\mu}}{\partial x^{\prime}_{\alpha}} \frac{\partial x_{\nu}}{\partial x^{\prime}_{\beta}} = g^{\prime}_{\mu \nu}$
So 
$g_{\mu \nu} A^{\mu} B^{\nu} = g^{\prime}_{\mu \nu} {A^\prime}^{\alpha} {B^{\prime}}^{\beta}$
And if the dot product is invariant so should be the angle between the two vectors.
So what's special about conformal transformations then?
Kindly tell me if I've done anything wrong here.
 A: Yes, all coordinate transformations preserve angles between vectors because coordinate transformations are simply changes in the way a particular manifold gets mapped into $\mathbb{R}^n$.  So a coordinate system for (an open subset of) a manifold $\mathcal{M}$ is a bijection $f: U\subseteq\mathcal{M}\to V\subseteq\mathbb{R}^n$.  Given two coordinate systems defined by $f$ and $g$ which cover the same patches of $\mathcal{M}$ and $\mathbb{R}^n$ then a coordinate transformation between them is just a map $g \circ f^{-1}: V\to V$.
Obviously such a thing can't change anything about the manifold or its metric: it's just changing how we represent it.  There are no coordinates in nature so changing coordinates can't change any physics, ever.
But a conformal transformation is not a coordinate transformation: it is a change in the metric itself.  In general such a change can alter angles between vectors, because we can choose whatever metric we like for $\mathcal{M}$ so long as it meets the requirements to be a metric.
A restricted sort of such transformations however, do preserve angles: ones where $\mathbf{g}'(p) = \alpha(p)\mathbf{g}(p)$, where $\mathbf{g}$, $\mathbf{g}'$ are the old and new metrics, $p\in\mathcal{M}$ and $\alpha(p)$ is a function of $p$.
You can, more-or-less equivalently, talk about conformal transformations as being diffeomorphisms between different manifolds such that the metrics are related by a scalar factor (or, really, the pulled-back metric is related by a scalar factor to the 'native' metric).  You can see that this is mostly the same thing but it allows you to think about conformal relationships between manifolds which are not obviously the same: Penrose diagrams are a good example of that.

Note I am talking here about conformal transformations (also called conformal mappings) as they're used in GR.  I think the other answer is really talking about conformal transformations between coordinate systems which is a different (although related) thing.  So I'm assuming here what the question is asking about and my assumption might be wrong.
A: Let $\mathcal{M}^d$ be a $d$-dimensional $(p,q)$ Riemannian manifold and let $f\colon \mathcal{M}^d\to\mathcal{M}^d$ be a coordinate transformation. Such transformation is said to be conformal if, by definition, 
$$g'_{\mu\nu}(x') = e^{\omega(x)}g_{\mu\nu}(x)$$
$x' = f(x)$ being the way the coordinates are affected by such transformation (to be intended component-wise, obviously). Working out the above definition one is led to the following set of possible transformations:
$$x'^{\mu} = x^{\mu} + a^{\mu}\qquad \textrm{translations}$$
$$x'^{\mu} = \lambda x^{\mu}, \quad \lambda\in\mathbb{R}\qquad \textrm{dilations}$$
$$x'^{\mu} = \Lambda^{\mu}_{\nu}x^{\nu}\quad \Lambda\in\textrm{SO}(p,q)\qquad \textrm{Lorentz}$$
$$x'^{\mu} = \frac{x^{\mu}-b^{\mu}x^2}{1-2b\cdot x +b^2 x^2} \qquad\textrm{special conformal}$$
the above are the only collection of transformation satisfying the original requirement for the metric to only change by a (positive) scaling factor. In standard terminology one says that conformal transformations preserve the angles as it is obviously so given the multiplicative pre-factor that cancels out when calculating the scalar products and then dividing by the norm of the original vectors.

Below I show that the dot product doesn't change by a co-ordinate transformation.

Well, the scalar product is, by definition of scalar, left unchanged by a change of coordinates, there is no need of calculating anything.
