Let $g$ and $g'$ be two $C^2$-smooth Riemannian metrics defined on neighborhoods $U$ and $U'$ of $0$ in $\mathbb R^2$, respectively. Suppose furthermore that the scalar curvature at the origin is $K$ under both metrics.

My question: Is there a coordinate transformation taking one metric to the other, such that they agree up to second derivatives at the origin? i.e., if $x : U \to U'$ is the transformation, we have

$g_{ij}' = g_{ab} ~x_i^a x_j^b$,

evaluating everything at $0$; there are similar equations for the first and second derivatives. Clearly this is false if the scalar curvatures aren't equal. I don't care what happens away from the origin.

In the excellent thread When is a Riemannian metric equivalent to the flat metric on $\mathbb R^n$?, Greg Kuperberg says:

If remember correctly, there is a more general result due to somebody, that any two Riemannian manifolds are locally isometric if and only if their curvature tensors are locally the "same".

If "local isometry" means that the metrics are equal on a neighborhood of the origin, then the metrics I have in mind are not locally isometric, since the only information I have is that their curvatures match at one point.