Group action on a condensed set and its orbit space

Question

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.

Answer 1

The answer is no (in general) for question 0 and no for question 2. In the edit of the question I have explained why $S \mapsto X(S)/G$ is not a condensed set in general.

In the following sense, the right generalization of the orbit space seems to be the categorical quotient (thanks to Echo's comment, I got to know this definition), and the categorical quotient of an action of $G$ on $X$ is the sheafification of $Y:S \mapsto X(S)/G$.

To see this, let's start by denoting $Y^+$ as the sheafification of $Y$. Let $\pi: X \rightarrow Y$ and $i: Y \rightarrow Y^+$ be the projection and the canonical morphism, respectively. For every $g \in G$, it is immediate that \begin{align*} i \circ \pi \circ g = i \circ \pi \end{align*} Now, let $Z$ be a condensed set with a map $q:X \rightarrow Z$, such that for each $g \in G$, we have \begin{align*} q \circ g = q \end{align*} Then, for each profinite set $S$, we have $q_S \circ g = q_S$. We define \begin{align*} \phi_S: Y(S) = X(S)/G &\longrightarrow Z(S) \\ \overline{x} &\longmapsto q_S (x) \end{align*} It is easy to check that $\phi_S$ is well-defined and that it defines a natural transformation $\phi:Y \rightarrow Z$. Also, note that $\phi$ is the only option to commute this diagram. Therefore, by the universal property of the sheafification, there is a unique map $Y^+ \rightarrow Z$ commuting the desired diagram.