I recently read that defining the action of $\mathbb{Z}$ on $\mathbb{R}$ by translation, i.e. $x \mapsto x+n$ for $x\in \mathbb{R}, n\in \mathbb{Z}$, gives that $\mathbb{R}/\mathbb{Z}\cong S^1$, and this gives that $\pi_1(\mathbb{R}/\mathbb{Z})\cong \mathbb{Z}$. I was wondering if there is any way to define a group action on $\mathbb{R}$ such that $\pi_1(\mathbb{R}/(\mathbb{Z}/n)) \cong \mathbb{Z}/n$.
EDIT: sorry if the question was confusing. I guess my main question is if there is any way to define a group action of $\mathbb{Z}/n$ on $\mathbb{R}$ in general. As a second question, would such an action satisfy $\pi_1(\mathbb{R}/(\mathbb{Z}/n)) \cong \mathbb{Z}/n$ where the quotient map $q:\mathbb{R}\to \mathbb{R}/(\mathbb{Z}/n)$ is a covering (in this case I believe the answer that Kenta S gave disproves this claim).
Consider an action of $\mathbb Z / n \mathbb Z$ on $\mathbb R$. I'll assume that this is an action in the topological category, i.e. an action by homeomorphisms, which is the most general situation that seems relevant to the tags of this post.
I'll lay out a proof that an action of $\mathbb Z / n \mathbb Z$ on $\mathbb R$ either is trivial and hence its quotient is just $\mathbb R$, or the action factors through a reflection action of $\mathbb Z/2\mathbb Z$ and so its quotient is homeomorphic to $[0,\infty)$.
Pick a generator of $\mathbb Z / n \mathbb Z$ and let $\alpha : \mathbb R \to \mathbb R$ denote the action of that generator.
By an argument applying the intermediate value theorem, one can prove that if $\alpha$ has no fixed points then it is either strictly increasing or strictly decreasing, implying that $\alpha$ has infinite order. [Side note: it follows furthermore that $\alpha$ is topologically conjugate to a translation homeomorphism $x \mapsto x+1$].
But $\alpha$ was assumed to have finite order, and so we obtain a contradiction.
We conclude that the set $F \subset \mathbb R$ of fixed ponts of $\alpha$ is nonempty, and it is a closed subset of $\mathbb R$.
Note that $\alpha$ acts on $\{-\infty,+\infty\}$, either fixing them both or transposing them, and we'll consider two separate cases.
Case 1: $\alpha$ fixes both $-\infty$ and $+\infty$. In this case we'll prove that $F=\mathbb R$, hence $\alpha$ is the identity map, hence the action is trivial.
For the proof, assuming that $F \ne \mathbb R$, consider a component $J$ of $\mathbb R - F$, which is an open interval with endpoints in the set $F \cup \{-\infty,+\infty\}$. The endpoints of $J$ are fixed by $\alpha$, and it follows that $\alpha(J)=J$. But $J$ is homeomorphic to $\mathbb R$, hence the restriction $\alpha \mid J$ has a fixed point not in the set $F$, which is a contradiction.
Note in this case that the quotient $\mathbb R / (\mathbb Z / n \mathbb Z)$ is just $\mathbb R$, whose fundamental group is trivial.
Case 2: $\alpha$ transposes $\{-\infty,+\infty\}$. It follows that $F$ is bounded, hence compact, and hence has a max and a min. Furthermore, $\alpha$ takes $\min(F)$ to $\max(F)$ and $\max(F)$ to $\min(F)$. But both $\min(F)$ and $\max(F)$ are fixed by $\alpha$, hence $\min(F) = \max(F)$ and so $F$ is a single point.
Applying Case 1 to $\alpha^2 = \alpha \circ \alpha$, it follows that $\alpha^2$ is the identity map. Hence $n$ is even and the action of $\mathbb Z / n \mathbb Z$ factors through an action of $\mathbb Z / 2 \mathbb Z$ whose fixed point set is the one-point set $F$. In this case the quotient is homeomorphic to $[0,\infty)$ whose fundamental group is trivial.
[Side note: There are indeed actions like this, because $\mathbb Z / 2 \mathbb Z$ can act as a reflection map $x \mapsto -x$ on $\mathbb R$. With just a bit more work, one can prove that every action of $\mathbb Z / 2 \mathbb Z$ is topologically conjugate to a reflection action. Hence if $\mathbb Z/n\mathbb Z$ acts nontrivially on $\mathbb R$ then $n$ is even, and the action is topologically conjugate to an action that factors through a reflection action of $\mathbb Z / 2 \mathbb Z$.]