I understand that hyperbolic spaces with different curvatures are not isometric. However, I wonder if this is also true for finite, discrete points on two hyperbolic spaces with different curvatures. Specifically when the number of points is n = 3 or n = 4, what's the conclusion?
Formally, consider two two-dimensional hyperbolic spaces $\mathbb{H}_2^{-K_1}, \mathbb{H}_2^{-K_2}$ with curvatures $-K_1 \neq -K_2$. Let ABC be a triangle on $\mathbb{H}_2^{-K_1}$, then is it possible to find A'B'C' on $\mathbb{H}_2^{-K_2}$, such that $\overline{AB} = \overline{A'B'}$, $\overline{AC} = \overline{A'C'}$ and $\overline{BC} = \overline{B'C'}$? Intuitively, mimicing what we would do in Euclidean space, one can first find two points $A',B'$ such that $\overline{AB} = \overline{A'B'}$; then draw two circles centered at $A', B'$ with radius $\overline{AC}, \overline{BC}$ respectively; and find the intersection point as $C'$. However, I'm not sure whether this procedure makes sense in hyperbolic space and seek for a theoretical answer here.
Also, if it's indeed possible for n = 3, then what about n = 4? n = 5?
Just in case someone would like to do analytical calculations, here are some basics for hyperbolic spaces with curvature $\neq -1$. Using Poincare disk model, the two hyperbolic spaces $\mathbb{H}_2^{-K_1}, \mathbb{H}_2^{-K_2}$ aforementioned can be characterized as same set of points: $\{z \in \mathbb{C}_2||z|\leq 1\}$; but with different distance metric: $d_1(z_1, z_2) = \frac{1}{\sqrt{K_1}}atanh(|\frac{z_1-z_2}{1-\bar{z_1}z_2}|), d_2(z_1, z_2) = \frac{1}{\sqrt{K_2}}atanh(|\frac{z_1-z_2}{1-\bar{z_1}z_2}|)$. (As you can see, the distance metric is only different up to scaling.)
All of the standard axioms of Euclid for planar geometry hold as well in $\mathbb H_2^{-K}$, except for the parallel axiom (or any of its equivalents). So your construction of triangles holds as well.
And that construction in fact shows that for a given numeric triple $(s_1,s_2,s_3)$ of positive numbers, that triple forms the side lengths of some triangle in $\mathbb H_2^{-K}$ if and only if it forms the side lengths of some Euclidean triangle, which holds if and only if the three "strict triangle inequalities" are satisfied: $$s_1 < s_2 + s_3 $$ $$s_2 < s_3 + s_1 $$ $$s_3 < s_1 + s_2 $$ If you look in detail at your geometric argument drawing those circles, those three inequalities are exactly what ensures that the two circles centered at $A',B'$ with radii $\overline{AC}$, $\overline{BC}$ actually do intersect at two points.