Let $X$ be a condensed set, and let $G$ be a (discrete) group. Suppose we have an action $G$ on $X$, which is a group morphism $a:G \rightarrow \mathrm{Aut}(X)$, where $\mathrm{Aut}(X)$ is the group of condensed isomorphisms. In this case, what would be an appropriate definition of the orbit space of $a$?

We may proceed as follows: For each profinite set $S$, we get an action $a_S: G \rightarrow \mathrm{Aut}(X(S))$, where $g \mapsto a(g)_S$, so we can define X/G as \begin{align*} (X/G)(S) = X(S)/G. \end{align*} Where $X(S)/G$ is the orbit space of $a_S$. This defines a presheaf on the category of profinite sets, but I think it might not be a condensed set without additional hypotheses.

$\textbf{Question 0:}$ Is $X/G$ a condensed set? (It is not, see Edit.)

$\textbf{Question 1:}$ If we have additional hypotheses, can we turn $X/G$ into a condensed set?

$\textbf{Question 2: (Main Question)}$ Is $X/G$ the appropriate definition of the orbit space of a condensed set?

$\textit{Edit:}$ I have just checked that even in very good cases, this definition of orbit space is not a condensed set. To see that, take $S^1 = \mathbb{R}/\mathbb{Z}$, $\alpha$ an irrational real number, and for each $n \in \mathbb{Z}$ and each profinite set $S$, we have \begin{align*} n: C(S, S^1) &\longrightarrow C(S, S^1) \\ f &\longmapsto f + n \alpha \end{align*} This defines an action in the condensed setting. Let $Y = S^1/\mathbb{Z}$. It is not true that $Y(S \sqcup T) = Y(S) \times Y(T)$. To see this, start with a map $f: S \sqcup T \rightarrow S^1$ and define $g(s) = f(s) + n\alpha$ for $s \in S$ and $g(t) = f(t) + m\alpha$ for $t \in T$ with $n \neq m$. This way, $f \neq g$ in $Y(S \sqcup T)$ but the map $\phi: Y(S \sqcup T) \rightarrow Y(S) \times Y(T)$ sends $f$ and $g$ to the same element. Thus, $\phi$ is not bijective, concluding the claim.