= Solution
Write a <proper Euclidean motion of Euclidean three-space> as $E(x)=Ax+b$, where $A\in SO(3)$. Then
$$
(E\circ\gamma)^{(j)}=A\gamma^{(j)}
\qquad(j\geq1).
$$
Because $A$ preserves norms, dot products, and cross products, the formulas
$$
\kappa=|T'|,
\qquad
\tau=\frac{\det(\gamma',\gamma'',\gamma''')}
{|\gamma'\times\gamma''|^2}
$$
show directly that curvature and torsion are unchanged by $E$.
For the shifted curve $\gamma_{s_0}(s)=\gamma(s-s_0)$, every derivative at $s$ equals the corresponding derivative of $\gamma$ at $s-s_0$. Consequently
$$
\kappa_{\gamma_{s_0}}(s)=\kappa_\gamma(s-s_0),
\qquad
\tau_{\gamma_{s_0}}(s)=\tau_\gamma(s-s_0).
$$
Both are therefore <pointwise Euclidean invariant of a curve>[pointwise Euclidean invariants] in the stated sense.
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