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Take the scalene triangle vertex set $T=\{(0,0),(1,0),(0,2)\}$. Its edge lengths are $1,2,\sqrt{5}$. 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. Only the identity preserves $T$.

For the general union of group translates, let $h\in G$. Left multiplication by $h$ permutes $G$, soTherefore $G\subseteq H$. This containment uses only the group action; the absence of symmetries of $T$ is not needed for the containment itself.

For a group of Euclidean isometries, every element preserves the displayed union of images of a set, because left multiplication permutes its group indices. Its full setwise isometry group can be larger even when $T$ itself has no nonidentity symmetry. Take $G$ to be the integer translations of the plane and $T=\{(0,0),(1,0),(0,2)\}$. By rigidity of a scalene triangle, $T$ has trivial stabilizer subgroup. Yet $S={\mathbb{Z}}^2$, whose quarter-turn about the origin is a symmetry outside the translation group.