I read that a conformal map is one that preserves the angles. I know nothing more about conformal maps. I don't know where to find a generalized definition of a conformal map, but I guess that if $f:U\to U$ is a map defined from and to a space endowed with an inner product $<\cdot,\cdot>$, as angles can be defined through the inner product, it should suffice for a conformal $f$ to have the following property:
$$<f(u),f(v)>=<u,v>\qquad\forall\ u,v\in U$$
Now, let's say that $U\equiv L^{2}(\mathbb{R})$ and $f\equiv\mathscr{F}$. Then
$$<\mathscr{F}f,\mathscr{F}g>=<f,g>\qquad\forall\ f,g\in U$$
by the Parseval-Plancherel theorem. So, is it the Fourier transform a conformal map on $L^{2}(\mathbb{R})$?
In general, a conformal map preserves angles (but not necessarily length).
In the case of an inner product space, since the angle between two vectors $v,w$ is defined to be the arccosine of thrle quantity $\frac{\langle v,w\rangle}{|v||w|}$, a conformal map $c:V\to W$ is one that preserves this quantity. In particular, since $|v| = \sqrt{\langle v,v\rangle}$, if $c$ preserves the inner product, it must preserve angles, hence is conformal.
Some more general context: a conformal map of Riemannian manifolds, $c:(M,g)\to (N,h)$, is one so that the angle between two tangent vectors $v_p,w_p\in T_pM$ is equal to the angle between their images $dc(v_p),dc(w_p)\in T_{c(p)}N$. As an exercise, you should prove that this is equivalent to the existence of a function $f:M\to\mathbb{R}$ such that $c^*h = fg.$