= Solution
Take the scalene triangle vertex <set> $T=\{(0,0),(1,0),(0,2)\}$. Its edge lengths are $1,2,\sqrt5$. Each vertex has a different unordered pair of incident edge lengths. A <Euclidean isometry> preserving the <set> must preserve these pairs, and therefore fixes every vertex.
To justify <rigidity of a scalene triangle> fully, <Euclidean distances> from an arbitrary plane point to these three <fixed points> determine that point uniquely: subtracting its squared <Euclidean distance> to the origin from the other two squared <Euclidean distances> determines its two coordinates. Thus an <isometry> fixing the three vertices fixes every plane point. \b[Only the identity preserves $T$.]
For the general union of <group> translates, let $h\in G$. Left multiplication by $h$ permutes $G$, so
$$
h(S)=\bigcup_{g\in G}hg(T)=\bigcup_{k\in G}k(T)=S.
$$
Therefore \b[$G\subseteq H$]. This containment uses only the <group action>; the absence of symmetries of $T$ is not needed for the containment itself.
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