This is not always possible. (This answer was worked out in collaboration with Raymond Cheng.)

**Example.** Let $X$ be the pseudocircle: it is the finite quotient of the unit circle $S^1 \subseteq \mathbb R^2$ where the open upper half segment is contracted to a point and the lower half segment is contracted to a point:

It is a four-point topological space with two closed points and two open points, and it is weakly homotopy equivalent to $S^1$. In particular, $\pi_1(X) = \mathbb Z$.

On the other hand, there are only finitely many possible quotients $X/{\sim}$, and therefore not every group $\mathbb Z/n\mathbb Z$ can occur as the fundamental group of $X/{\sim}$. $\square$

**Remark.** We can actually go through all possible quotients. If $\sim$ is a nontrivial equivalence relation, then $X/{\sim}$ is connected and has fewer than $4$ points. In fact, it is always contractible:

- Any one-point space $Y$ is contractible.
- Any connected two-point space $Y$ is either Sierpiński space or indiscrete, both of which are contractible¹.
- If $|X/{\sim}| = 3$, then we have either:
- identified two open points to give the 'pseudo-closed-interval', which is contractible¹;
- identified two closed points to give a 'pseudo-open-interval', which is contractible¹;
- identified a closed point with an open point, giving a three-point space where one point is open, one point is closed, and one point is neither. This is the finite $T_0$ space corresponding to a total order, and it is contractible¹.

¹For example, consider the 'pseudo-closed-interval' $Y$:

Labelling the points on $Y$ by $0, \varepsilon$, and $1$, we can contract $Y$ to a point by
\begin{align*}
\Phi \colon Y \times [0,1] &\to Y \\
(y,t) &\mapsto \left\{\begin{array}{ll} y, & t = 0 \\ 0, & y = 0, \\ \varepsilon, & 0 < t < 1 \text{ and } y \neq 0, \\ 0, & t = 1. \end{array}\right.
\end{align*}
The inverse image of $0$ is the closed set $(Y \times \{1\}) \cup (\{0\} \times [0,1])$, and the inverse image of $1$ is the closed set $\{(1,0)\}$, proving that $\Phi$ is continuous. Hence, $\Phi$ is a homotopy from $\operatorname{id}_Y$ to the constant map $0$, showing that $Y$ is contractible. The proofs for the other spaces are similar.