If you look at the book Conformal Field Theory (by Philippe Francesco, Pierre Mathieu and David Senechal) or the lecture notes Applied Conformal Field Theory (by Paul Ginsparg), and many other places: The conformal transformation is defined as the subset of coordinate transformations under which the metric changes as follows, $$ g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x) $$ under a coordinate change $x'=x'(x)$ where, $$ g_{\mu \nu}'(x')=\frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}} g_{\alpha \beta}(x). $$ But I think the first equation should be: $$ g_{\mu \nu}'(x') = \Omega(x') g_{\mu \nu}(x'). $$ Although this looks like a typo, but it is consistant in many lectures/books. So I am little confused. Am I thinking it wrong? Books/Lectures are probably thinking that in the end of the transformation, they are relabeling $x'$ as $x$. But then that should be explicitly mentioned.
Derivation of conformal killing vector equation using book's version of the equation: Consider an infinitesimal change, $$ x'^{\mu} = x^{\mu} + \epsilon \xi^{\mu}(x) $$ and, $$ \Omega(x) = 1+\epsilon \omega(x) $$ Then the equation according to the books: $$ g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x) $$ gives, $$ g_{\mu \nu}'(x') = g_{\mu \nu}(x) + \epsilon \omega(x) g_{\mu \nu}(x) $$ Also we can expand, $$ g_{\mu \nu}(x') = g_{\mu \nu}(x) + \epsilon \xi^{\alpha} \partial_{\alpha} g_{\mu \nu}(x) $$ The Lie derivative of the metric, $$ \mathcal{L}g=\lim_{\epsilon \to 0} \left( {g'_{\mu\nu}(x')-g_{\mu\nu}(x') \over \epsilon} \right) = \omega(x) g_{\mu \nu}(x) -\xi^{\alpha} \partial_{\alpha} g_{\mu \nu}(x) $$ Which implies, $$ -(\xi_{\mu;\nu}+\xi_{\nu;\mu})=\omega(x) g_{\mu \nu}(x) -\xi^{\alpha} \partial_{\alpha} g_{\mu \nu}(x) $$ as the killing vector equation. Where as if you consider the equation: $$ g_{\mu \nu}'(x') = \Omega(x') g_{\mu \nu}(x') $$ Then you get following similar steps, $$ -(\xi_{\mu;\nu}+\xi_{\nu;\mu})=\omega(x) g_{\mu \nu}(x) $$ Then which Killing vector equation is correct?