= Solution
At $p\in D$, each of the given ambient rotations fixes $p$, rotates its tangent <plane> and preserves the local surface. Naturality of the <shape operator> under that rotation makes $S_p$ commute with every planar rotation. A symmetric real two-by-two operator commuting with all rotations is scalar, so every point of $D$ is an <umbilical point>. The trace-free part of the <shape operator> is continuous. Density therefore makes it zero everywhere on $S$.
Locally write $S=\kappa I$. The <Codazzi equation> gives $(X\kappa)Y=(Y\kappa)X$ for tangent fields $X,Y$; choosing independent fields proves $d\kappa=0$. Thus the surface is locally a <plane> if $\kappa=0$, or has constant spherical center
$$
p+\frac{N(p)}\kappa,
$$
since its differential is $I-S/\kappa=0$. A change of normal reverses both $N$ and $\kappa$, leaving this center unchanged, so these local descriptions patch across the connected surface. In the flat case the whole surface lies in one <plane>, where a <regular surface> without boundary is open; a nonempty such set cannot also be compact and closed. Hence the flat case is excluded. In the other case $S$ lies on one <sphere> of radius $1/|\kappa|$. Its inclusion in that <sphere> is locally open, since both have dimension two, and its <compactness> makes the image closed. <Connectedness> of the <sphere> forces the image to be the entire <sphere>. Therefore \b[$S$ is a <sphere>].
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