This question is motivated by Problem 10.2 of "Geometry from a Differentiable Viewpoint" by McCleary.
Question: Let $S\subset\mathbb{R}^3$ be a regular orientable surface, $N:S\longrightarrow\mathbb{S}^2$ be the corresponding Gauss map, and $\rho:\mathbb{R}^3\longrightarrow\mathbb{R}^3\in O(3)$ be an orthogonal linear transformation such that $\rho(S)=S$ (in other words, the surface $S$ is symmetric with respect to the rigid motion $\rho$).
Then, can we say $N\circ\rho\vert_S=\rho\vert_{\mathbb{S}^2}\circ N$?
In other words, I am wondering if the surface is symmetric, then the corresponding Gauss map is also symmetric. Intuitively I think it must be true, but I could not construct a rigorous proof. This kind of question must be classical, is there any reference dealing with this question?
Yes, the Gauss map is equivariant with respect to rigid motions in your sense, provided we're careful with orientation.
If $p$ is a point of $S$, and if $u$ and $v$ comprise a basis of the tangent plane $T_{p}S$ such that $N(p) = u \times v$, then $\rho(u)$ and $\rho(v)$ comprise a basis for the tangent plane $T_{\rho(p)}S$ because $S$ is invariant under $\rho$. Consequently, $$ \rho(N(p)) = \rho (u \times v) = \det(\rho)\rho(u) \times \rho(v) $$ is a unit vector orthogonal to $T_{\rho(p)}S$. Since $S$ is oriented and the ordered triple $(\det(\rho)\rho(u), \rho(v), N(\rho(p)))$ has the same orientation as $(u, v, N(p))$, we have $$ \rho(N(p)) = N(\rho(p)). $$