Fundamental group via deck transformations considering rotation on sphere Let $Z_m$ act on $S^1$ by multiplication with $e^{2\pi ki/m}$ for $k \in Z_m$. Let $X = S^1 / Z_m$ be the orbit space of this action. Then we have a universal cover $q:S^1 \rightarrow X$ given by the canonical projection. To determine the fundamental group of $X$ we can consider the group of deck transformations $Aut(q)$ of $q$, since we know that $\Pi_1(X) = Aut(q)$. 
$Aut(q)$ is by definition the group of all homeomorphisms $f: S^1 \rightarrow S^1$ such that $qf=q$. We see that all rotations by degree $e^{2\pi ki/m}$ for $k \in Z_m$ are such homeos. But why are there no other ones? 
 A: $\pi_1(X)$ is not meant to be equal to ${\rm Aut}(q)$. That would be true if the covering was a universal cover, which is not the case here.
[An example of a universal cover of $X$ would be the map $q_{\rm univ} : \mathbb R \to X$, sending $x \mapsto x {\rm \ mod \ } \tfrac{2\pi}{m}$. This is a universal cover because $\mathbb R$ is simply connected. The deck transformation group for this universal cover is ${\rm Aut}(q_{\rm univ}) = \mathbb Z$. Since $X$ itself is a small circle, $\pi_1(X) = \mathbb Z$, which agrees with ${\rm Aut}(q_{\rm univ})$.]
As for your original covering $q : S^1 \to X$, we still have the following useful result (Hatcher 1.39): If $H : = q_\star (\pi_1(S^1)) $ is the image of $\pi_1(S^1)$ under the group homomorphism $q_\star : \pi_1(S^1) \to \pi_1(X)$, then ${\rm Aut}(q)$ is isomorphic to $N(H) / H$, where $N(H)$ is the normaliser of $q_\star (\pi_1(S^1))$ in $\pi_1(X)$. Now $\pi_1(X) = \mathbb Z$, and $H$ is the subgroup $m\mathbb Z$, which is a normal subgroup, so the theorem tells us that ${\rm Aut}(q) = \mathbb Z_m$. And this agrees perfectly with your counting.
Finally, I'll address the question in your comment: Can we see directly that there are no more than $m$ deck transformations for $q$? The answer is yes, because $q$ is an $m$-sheeted covering of $X$ (meaning that the preimage of $q^{-1}(x)$ for any point $x \in X$ consists of $m $ points. Given a point $x \in X$, a deck transformation $f$ is uniquely determined by where it sends $x$. But $f$ can only send $x$ to one of the points in $q^{-1}(x)$, and there are $m$ such points. Hence there can be at most $m$ deck transformations.
