How group action of $G$ on set $\Omega$ acts on the vector space induced by $\Omega$, $\mathcal{X}(\Omega)$?

Question

A left group action of $G$ on a set $\Omega$ is mapping $G \times \Omega \to \Omega$. We define a space of $\mathcal{C}$ valued signals as $\mathcal{X}(\Omega.\mathcal{C}) = \{x:\Omega \to \mathcal{C}\}$ has a vector space property. As a typical illustration, take $\Omega = \mathbb{Z}_n \times \mathbb{Z}_n$ as a two dimensional $n \times n$ grids, $x$ is RGB image(i.e. a signal $x: \Omega \to \mathbb{R}^{3}$). Now if we want to obtain the group action of $G$ on the space $\mathcal{X}(\Omega)$ we have: \begin{align} (g \cdot x)(u) = x(g^{-1}u) \tag{1} \end{align} where $g \in G, x \in \mathcal{X}, u \in \Omega$. I got the equation $1$ from the paper, Geometric Deep Learning Grids, Groups, Graphs, Geodesics, and Gauges (page 14 equation 3). The authors of this paper did not give any intuitive understanding of the equation $1$ also did not mention where to find the proof. Can anyone please help me to understand how this formula actually work?

Answer 1

They've defined a new group action $G \times \mathcal{X}(\Omega) \to \mathcal{X}(\Omega)$ using the old group action $G \times \Omega \to \Omega$.

If you have $x \in \mathcal{X}(\Omega)$, they define $g \cdot x$ as a signal $u \mapsto x(g^{-1} u)$ which belongs to $\mathcal{X}(\Omega)$. (I guess one should check that this satisfies the conditions that a group action must satisfy, like $(gh) \cdot x = g \cdot (h \cdot x)$.)