As I know the geodesic curvature $$ \kappa_g = \sqrt{\det~g} \begin{vmatrix} \frac{du^1}{ds} & \frac{d^2u^1}{ds^2} + \Gamma^1_{\alpha\beta} \frac{du^\alpha}{ds} \frac{du^\beta}{ds} \\ \frac{du^2}{ds} & \frac{d^2u^2}{ds^2} + \Gamma^2_{\alpha\beta} \frac{du^\alpha}{ds} \frac{du^\beta}{ds} \end{vmatrix}, $$ where $g$ is the metric tensor, $\Gamma^v_{\alpha\beta}$ is the Christoffel symbols of the second kind.
And the first fundamental form of the surface $I = (du^1, du^2) g (du^1, du^2)^T$. I think $I$ is invariant under isometric transformations but not the metric tensor $g$. So why $\kappa_g$ is invariant under isometric transformations?
This is a well known sequence of isometric transformations of the (left-handed) Helicoid into the Catenoid back into the (right_handed) Helicoid. I've put in examples of the principal lines of curvature, which are both geodesics . The drawing isn't perfect , the surfaces lying on the right and left sides are missing a red straight line ( a geodesic) down the center line of the helicoid. (Compare the central line on the Catenoid -- a circle ), from edge on the blue lines of the Helicoid are also straight lines. Incidently the drawing is quite large, you can get a better idea of it by taking it down off of your screen.
A reference for $\kappa_g$ is "Lectures on Classical Differential Geometry" by Dirk J. Struik ( Chapter Four , Geometry on the Surface , pp. 128 ). It is shown that geodesic curvature depends only on E,F,G hence it is an invariant under isometric transformations, or a bending invariant as it is called in the text.