Consider the gyroid surface: $$ \mathcal{G} = \{(x, y, z) \in \Bbb{R}^3 \ | \ f(x, y, z) = 0 \} $$ where $$ f(x, y, z) = \sin(x+y) + \sin(x-y) + \sin(y+z) + \sin(y-z) + \sin(x + z) + \sin(x-z). $$ Then $\mathcal{G}$ divides $\Bbb{R}^3$ in two connected regions, given by $\{f < 0\}$ and $\{f > 0\}$.
My question is:
What is a rigid motion that interchanges the two regions above?
After trying some translations by $\pi$, which did not work, I came to believe such a rigid motion is given by some rotation, but I am unable to find out what exactly.
Thanks in advance.
A direction of research, not a full answer.
Your issue is to find in a first step the invariance group of this surface (group of rigid motions preserving it). Those exchanging the two sides are order-2 elements ($T^2=Id$). Caution: the reciprocal isn't necessarily true.
Let us take the more compact equation :
$$\cos(x)\sin(y)+\cos(y)\sin(z)+\cos(z)\sin(x)=0\tag{1}$$
Remark : we don't take into account translations
$$x \to x+2k_x\pi, y \to y+2k_y\pi, z \to z+2k_y\pi,$$
because they shouldn't exchange interior and exterior (although it has to be verified...)
It suffices to consider symmetries.
(I don't consider here for example symmetries with respect to a line such as $(x,y,z)\mapsto (-x,-y,z)$ which can be considered as $\pi$ rotation around axis this line, knowing that, in general, rotations will be treated as compositions of symmetries with respect to planes).
$$\begin{pmatrix}x\\ y\\ z \end{pmatrix} = S_N \begin{pmatrix}X\\ Y\\ Z\end{pmatrix} \ \ \text{ with} \ \ S_N:=I_3-2N^TN = \begin{pmatrix}(1-2a^2)&-2ab&-2ac\\ -2ab&(1-2b^2)&-2bc\\ -2ac&-2bc&(1-2c^2)\end{pmatrix} \tag{2}$$
(where $N=(a,b,c)^T$ is a unit normal vector to the plane of symmetry ; we can check that trace($S_N)= 3-2(a^2+b^2+c^2)=1$ which is the sum of the eigenvalues $1,1,-1$ of $S_N$).
Then plug expressions (2) into (1) obtaining an expression that has to be identical to the initial expression, whatever the values of variables $(x,y,z)$ (or $X,Y,Z$).
Connected : this.