 Solution

Let $M_t(x)=R(t)x+T(t)$ be the given family. Composing throughout with $M_0^{-1}$ lets us assume $M_0$ is the identity; the family remains regular and preserves $\gamma$. Differentiating at $t=0$ produces a nonzero infinitesimal proper Euclidean motionFor every $x\in \gamma$, the path $t\mapsto M_t(x)$ lies in $\gamma$, so $V(x)$ is tangent to $\gamma$. The flow of this complete vector field consequently preserves $\gamma$ for all times.

The orbit of a one-parameter proper Euclidean motion is a line if $a=0$, and otherwise a helix about an axis parallel to $a$. Since $\gamma$ is compact, it cannot contain an entire line or a helix with nonzero axial pitch. Thus the flow is a pure rotation about an axis. The curve cannot lie entirely in the fixed axis, because no subset of a line is a simple closed curve. Choose a point off the axis. Its orbit is a circle contained in $\gamma$; as a one-dimensional orbit it is open in the connected one-dimensional manifold $\gamma$, and it is closed by compactness. It must therefore equal the whole curve. Hence the continuous rigid-motion symmetry of a simple closed space curve gives