For a given natural number \(n > 1\), the cyclic group \(\mathbb{Z} / n \mathbb{Z}\) consists of the residue classes of integers, modulo \(n\). If we use the notation \([a]_n\) to refer to the residue class of \(a \in \mathbb{Z}\), modulo \(n\), then the elements of \(\mathbb{Z} / n \mathbb{Z}\) are \[ \{[0]_n, [1]_n, \dots, [n-1]_n\} \] The group operation (which we call "+") is addition, modulo \(n\), so that \([a]_n + [b]_n := [a+b]_n\).
Every integer \(c \in \mathbb{Z}\) gives rise to a homomorphism from the group to itself (a.k.a. an endomorphism) \[ \phi_c : \mathbb{Z} / n \mathbb{Z} \longrightarrow \mathbb{Z} / n \mathbb{Z} \] defined by \(\phi_c([a]_n) := [ca]_n\).
We can visualize these endomorphisms by identifying \(\mathbb{Z} / n \mathbb{Z}\) with the \(n\)th roots of unity \[ U_n := \{e^{2\pi i / k} | i=0,\dots,n-1\} \subset \mathbb{C} \] which is a subgroup of the unit circle group \(S^1 \subset \mathbb{C}\). To do this, define \(\psi: \mathbb{Z} / n \mathbb{Z} \rightarrow U_n\) by \(\psi([a]_n) = e^{2\pi i a / n}\). One can check that this is a homomorphism, and that it is injective. By construction the image is \(U_n\), so we have an isomorphism \(\mathbb{Z} / n \mathbb{Z} \simeq U_n\).
Under this identification, an endomorphism \(\phi_c\) becomes the mapping \[ e^{2\pi i k/n} \longmapsto e^{2\pi i ck / n} \] We can visualize the endomorophism by drawing a line segment from each input point to the corresponding output point.
The visualization below iterpolates between different endomorphisms \(\phi_c\) as the parameter \(c\) varies over time. The group has \(n=512\). There is also a larger version, more suited to desktop browsers..