General Relativity in a Differential Geometry concept I know that in $\mathbb{R}^2$ we can define the curvature of a parametrized curve $\textbf{x}(t)=\bigl( x(t), y(t)\bigr)$ as
$$ \kappa(t) = \dfrac{\text{det}(\textbf{x}',\textbf{x}'')}{||\textbf{x}'||^3}.$$
If I use a change of parameter $\overline{t}=\phi(t)$ we can reparametrize the curvature
$$ \boxed{\overline{\kappa}(\overline{t}) =\kappa(t)\equiv \kappa\bigl(\phi(\overline{t})\bigr)}\tag{1}.$$
Equation (1) is the definition of a scalar. The curvature is a scalar magnitude.

My question is: 1. Why is the meaning of the next equation, very similar
  to equation (1)? 2. In particular, what functions verify the next
  equations? $$ \overline{\kappa}(t)=\kappa(t) .\tag{2}$$
  My teacher says that the solution is in a General Relativity book, usually in the 7th or 8th chapter. 3. So what is the General Relativity concept behind the equation (2)?

My opinion: It could be the curvature $\kappa: \mathbb{R} \rightarrow \mathbb{R}$ of a geodesic curve in a surface. The curvature of all curves verify the equation (1), in addition if the curve is a geodesic also verify the equation (2). Indeed, this concept is related to General Relativity.
Note that equation (2) is not the same that the equation (1).
I asked the same question in Mysterious property of the curvature of a curve in $\mathbb{R}^2$
 A: In $(1)$, the new curvature $\bar{\kappa}$ and the old curvature $\kappa$ are evaluated at different corresponding values of the coordinate. Say you have your curve $\mathbf{x}(t)$, which you used to calculate $\kappa(t)$, and you make a reparametrization $\bar{t} = 2t$. The first equation says that if you use this new parametrization to calculate $\bar\kappa$ at $\bar t = 2$, you'll get the same value as if you calculated $\kappa$ at $t=1$, because $t=1$ is the same point of the curve as $\bar t = 2$. This is pretty straightforward: if you change your parameter, the same point of the curve will have different values of the parameter, and the curvature is a scalar: it just depends on where you are on the curve, not on your coordinate.
In $(2)$, notice how it says $t$ in both sides. Now we're saying that the curvature won't change if you calculate it using both parametrizations but using the same value of the coordinate, not corresponding values like earlier. To reuse our example, this equation says that $\bar \kappa$ evaluated at $2$ should give the same value as $\kappa$ evaluated at $2$, not at $1$ like before.
The first important thing to say is that this is not necessarily true. Equation $(1)$ says that the curvature is a scalar and it is always true, but equation $(2)$ is a property that may or may not hold, depending on your curve and the transformation. This is similar but not exactly the same as an isometry,  meaning a rigid movement: the curve has some symmetry which allows us to change coordinates while the curvature looks the same. The same concept is used in GR, just in more dimensions.

Let's do an example. Let's take a parabola as our curve: $x(t) = t, y(t) = t^2$. The curvature ends up being
$$\kappa(t) = \frac{2}{(1+4t^2)^{3/2}}.$$
For example, $\kappa(1) \approx 0.18$. If we change to $\bar t = 2t$, we get a new function
$$\bar \kappa (\bar t) = \frac{2}{(1+\bar t^2)^{3/2}}.$$
Notice that as functions of their arguments, $\bar \kappa$ and $\kappa$ are different, but if you evaluate them at arguments such that $\bar t = 2t$, they give the same result: $\bar \kappa(4) = 0.18 = \kappa(2)$. However, there is no relation between $\bar \kappa(2)$ and $\kappa(2)$. But if we define $t' = -t$, the curvature is the same function of $t'$ as it was of $t$. Now we will have identities like $\kappa'(2) = \kappa(2)$. This is because the transformation we did is a symmetry of the parabola: it corresponds to flipping it around its axis. If our curve was a circle, then any transformation would preserve the curvature, because the curvature is constant.
