= Solution
Let
$$
\beta(s)=\alpha(\pm s+\sigma_0).
$$
This is another unit-speed <embedded curve>. The assumed identities say that $\alpha$ and $\beta$ have the same positive curvature and the same torsion. By the uniqueness part of the <fundamental theorem of regular space curves>, there is a proper Euclidean motion $\phi$ such that
$$
\phi(\alpha(s))=\beta(s)=\alpha(\pm s+\sigma_0).
$$
Since $s\mapsto\pm s+\sigma_0$ is a bijection of $\mathbb R$, this motion preserves $\alpha(\mathbb R)$.
It is nontrivial. For the plus sign, $\sigma_0\ne0$ and $\phi=\operatorname{id}$ would imply $\alpha(s)=\alpha(s+\sigma_0)$, contradicting embeddedness. For the minus sign, the identity would imply $\alpha(s)=\alpha(-s+\sigma_0)$ for every $s$, and injectivity would force $s=-s+\sigma_0$ for every $s$, which is impossible. Thus the required nontrivial proper Euclidean motion exists.
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