If I have a metric $g$ and $h=ag$ for some positive constant $a$, how does the Gauss curvature of $h$ ($K_h$) relate to the Gauss curvature of $g$ ($K_g$). I believe it should be $a^{-1}$ but I am not sure why (I'm somewhat new to this). Could someone let me know if this is correct (and maybe a very brief reason why). I'm sorry if this is a simple question, but I would like to use it for a paper I am working on.
If it is helpful, I am working with metrics on $S^2$ and I have conformal metrics $g_2=e^{2w}g_1$ and I would like to conclude that $K_{h}=e^{-2a} K_{g_2}$ where $h=e^{2w+2a}g_1=e^{2a}g_2$.
Since the Gaussian curvature is the product of two geodesic curvatures, which scale as $\ell^{-1}$ (where $\ell$ is length as measured by the metric), it scales as $\ell^{-2}$. The metric itself is of scale $\ell^2$, so you're right that $h=ag$ implies $K_h = a^{-1}K_g.$
If you want a computational proof, you could start with one of the explicit formulae for $K_g$ in terms of $g$ and it should fall out pretty quickly.