= Solution
Write $\xi=V_1$, $\eta=\phi$, and $u_j=d^ju/dx^j$. On the <jet space of a scalar ordinary differential equation>, define
$$
\eta^{(0)}=\eta,
\qquad
\eta^{(j)}=D_x\eta^{(j-1)}-u_jD_x\xi,
$$
where the <total derivative operator> is
$$
D_x=\partial_x+u_1\partial_u+u_2\partial_{u_1}+\cdots.
$$
The order-$N$ <prolongation of a vector field> is
$$
\boxed{
\operatorname{pr}^{(N)}V
=\xi\partial_x+\eta\partial_u
+\sum_{j=1}^N\eta^{(j)}\partial_{u_j}}.
$$
In the coordinates $(x,u)$, the given action of the <special orthogonal group> is
$$
\widetilde x=x\cos s+u\sin s,
\qquad
\widetilde u=u\cos s-x\sin s.
$$
Differentiating at $s=0$ gives the <infinitesimal generator of a Lie point symmetry>
$$
V=u\partial_x-x\partial_u,
$$
so $\xi=u$ and $\eta=-x$. The recursive formula gives
$$
\eta^{(1)}=D_x(-x)-u_xD_xu=-1-u_x^2
$$
and
$$
\eta^{(2)}=D_x(-1-u_x^2)-u_{xx}D_xu
=-3u_xu_{xx}.
$$
Therefore the <second prolongation of the rotation generator for plane graphs> is
$$
\boxed{
\operatorname{pr}^{(2)}V
=u\partial_x-x\partial_u
-(1+u_x^2)\partial_{u_x}
-3u_xu_{xx}\partial_{u_{xx}}}.
$$
For $\Delta_0=u_{xx}$,
$$
\operatorname{pr}^{(2)}V(\Delta_0)=-3u_xu_{xx},
$$
which vanishes when $\Delta_0=0$. For
$$
\Delta_1=u_{xx}-(1+u_x^2)^{3/2},
$$
we obtain
$$
\begin{aligned}
\operatorname{pr}^{(2)}V(\Delta_1)
&=-3u_xu_{xx}
-3u_x(1+u_x^2)^{1/2}[-(1+u_x^2)]\\
&=-3u_x\Delta_1,
\end{aligned}
$$
which likewise vanishes on the equation. This is the infinitesimal invariance criterion for both equations.
Geometrically, $u_{xx}=0$ describes straight lines. The <signed curvature of a plane graph> is
$$
\kappa=\frac{u_{xx}}{(1+u_x^2)^{3/2}},
$$
so the second equation says $\kappa=1$ and describes consistently oriented arcs of unit circles. <rotation matrix>[Rotation] preserves straight lines, circles, and signed curvature, explaining both invariances.
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