Suppose $A \subset \mathbb{T}_2$ is a measurable set, where $\mathbb{T}_2$ is the torus group.
For a fixed $n$, define the function $f_A:\mathbb{T}_2^n \rightarrow \mathbb{R}$ as
$$f_A(x_1, ..., x_n) = \begin{cases} 1 \ \ \ \text{ if } x_1 \in A, x_2 \in A ... ,x_n \in A \\ 0\ \ \ \text{ otherwise}\end{cases}.$$
Now with the diagonal action of $\mathbb{T}_2$ on $\mathbb{T}_2^n$ (written multiplicatively), define the function $g_A:\mathbb{T}_2^n\rightarrow \mathbb{R}$ as
$$g_A(x) = \int_{\mathbb{T}_2}f_A(gx) dg $$
Is the function $g_A$ continous? It is smooth? What if I consider general a Lie group $G$ acting on a manifold $M$, and consider $A \subset M$ to make a similar construction?
Let $f_n$ be a sequence of continuous functions converging to $f_A$ (one can think of many such constructions).
Then $f_n$ are uniformly continuous because the domain is compact. One can easily prove that $g_n: x \mapsto \int_{g \in \mathbb{T}_2} f_n(g x) dx$ uniformly converge to $g_A$, and therefore $g_A$ must be continuous. A similar proof should generalize to the case of a compact group acting on a (compact?) manifold.
Smoothness is too much to ask for. We can take $n=2$ and $A$ to be a horizontal strip. Then the value of $g_A(x,y)$ is just a piecewise linear function only depending on the vertical distance of the points $x,y \in \mathbb{T}_2$