$$\newcommand{\zphom}{\mathbf{Grp}(\mathbb{Z}_p, \dash)} \newcommand{\zphoma}[1]{\mathbf{Grp}(\mathbb{Z}_p, #1)}$$

For $p$ prime, let $U_p : \mathbf{Grp} \rightarrow \mathbf{Set}$ be a functor sending a a group $G$ to its set of elements having order $1$ or $p$. For a group homomorphism $f : G \rightarrow H$, simply let $U_p(f)$ be the restriction of $f$ to $U_p(G)$ -- this is well defined since $p$ is prime, and thus $f$ necessarily sends an element of order $p$ to an element of order $1$ or $p$.

Theorem:

$U_p$ is representable. In particular, $U_p$ is isomorophic to $\mathbf{Grp}(\mathbb{Z}_p, \dash)$.

Proof: Define a natural transformation $\alpha : U_p \rightarrow \zphom$ as follows. For a group $G$, let $\alpha_G : U_p(G) \rightarrow \zphoma{G}$ send an element $g$ of order $1$ or $p$ to the unique homomorphism $\varphi : \mathbb{Z}_p \rightarrow G$ such that $\varphi(1) = g$. In other words, $\varphi$ is the map $n \mapsto g^n$. We now check naturality. Suppose $f : G \rightarrow H$ is a homomorphism. We must have $\alpha_H \circ U_p(f) = \zphoma{f} \circ \alpha_G : U_p(G) \rightarrow \zphoma{H}$. Recall that $\zphoma{f}$ is just left-composition by $f$. Take a $g$ in $U_p(G)$. On the right side, we have $\alpha_G(g) = n \mapsto g^n$, then left-composing with $f$ yields $f \circ (n \mapsto g^n) = n \mapsto f(g^n) = n \mapsto f(g)^n$. On the left, we have $U_p(f)(g) = f(g)$. Then, $\alpha_H(f(g)) = n \mapsto f(g)^n$. So both sides yield the same map $\mathbb{Z}_p \rightarrow H$, verifying that $\alpha$ is natural. So $U_p \cong \zphoma{\dash}$.