This is sort of an answer and sort of not. I'll let you be the judge:

Suppose you formulate the question, not in terms of 'motion' (which you left vague) but terms of 'freely copying' a triangle $T$, as follows:

Let $(S,g)$ be a Riemannian surface, and let $T$ be a triangle, i.e., a triple of points $(p_1,p_2,p_3)$ on the surface together with three geodesic segments $\gamma_{ij}=\gamma_{ji}$ for $i\not=j$ where $p_i$ and $p_j$ are the endpoints of $\gamma_{ij}$. Let $\ell_{ij}>0$ be the length of the geodesic segment $\gamma_{ij}$.

Now, suppose that, side-angle-side holds for this specific $T$ in the following sense: Whenever $(q_1,q_2,q_3)$ are three points of $S$ and $\eta_{12}$, respectively $\eta_{13}$, are geodesic segments of length $\ell_{12}$, respectively $\ell_{13}$, with endpoints $q_1$ and $q_2$, respectively $q_1$ and $q_3$, so that the angle between these geodesic segements at $q_1$ is the same as the angle between the geodesic segments $\gamma_{12}$ and $\gamma_{13}$ at $p_1$, then there exists a geodesic segment $\eta_{23}$ of length $\ell_{23}$ with endpoints $q_2$ and $q_3$, such that for all $i,j,k$ distinct, the angle between $\eta_{ij}$ and $\eta_{ik}$ is the same as the angle between $\gamma_{ij}$ and $\gamma_{ik}$.

Does it then follow that $S$ has constant Gauss curvature?

The answer is 'no', even if $S$ is a sphere: Let $(S,g)$ be a Zoll sphere all of whose geodesics are closed and of length $L$. Fix a point $p$ and let $T$ be a triangle with vertices $(p_1,p_2,p_3) = (p,p,p)$ and let the sides be any three geodesic segments $\gamma_{ij}$ of length $L$ with endpoints $p_i=p$. This 'triangle' can be copied to any triangle $T'$ with vertices $q_i=q$ by choosing the geodesic segments $\eta_{ij}$ so that the angles between sides are equal to the angles between the corresponding sides of $T$.

congruentto $T$, i.e., that there should be 'enough' of them in an appropriate sense. $\endgroup$ageodesic triangle", not "Let $T$ beanygeodesic triangle". Ifanygeodesic triangle can be copied without 'distortion', then, sure, the Gauss curvature has to be constant. This is an easy consequence of geodesic normal coordinates. I thought you were asking about the harder problem of knowing only that you can copy a specific 'congruence class' of geodesic triangles without distortion. $\endgroup$3more comments