= Solution
Let $R_t$ be the horizontal <rotation matrix> through <angle> $t$. Define the <screw motion>
$$
\boxed{\Phi_t(x,y,z)=(x\cos t-y\sin t,\ x\sin t+y\cos t,\ z+t).}
$$
Its linear part is an orthogonal <matrix> with <determinant> one, and its remaining part is a translation. Thus it is a <proper Euclidean motion of Euclidean three-space>, with $\Phi_s\Phi_t=\Phi_{s+t}$.
In horizontal-vector notation the given <parametric surface> satisfies $\mathbf r(u,v)=(R_v(1+\cos u,\sin u),v)$. Therefore
$$
\Phi_t(\mathbf r(u,v))=\mathbf r(u,v+t).
$$
The defining circle in each horizontal plane is entirely parametrized by $u$, so $\mathbf r$ covers $X$. The displayed equality and $\Phi_{-t}=\Phi_t^{-1}$ prove $\Phi_t(X)=X$. This <one-parameter subgroup> preserves the <dot products> and <cross products> defining $I$, $II$ and the chosen <unit normal>. Consequently their coordinate coefficients, and the <Gaussian curvature>, depend only on $u$.
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