Consider a map $\phi$ between two groups, $G$ and $H$, such that $\forall n \in \mathbb{Z}$ and $\forall a \in G$ we have $\phi(a^n)=\phi(a)^n$
Trivially, the identity of $G$ maps to the identity of $H$, but what else can we say about this type of map? In short, I'm looking for a structure-preserving map that preserves exponentiation.
$\phi\Bigg|_{\langle a \rangle}:\langle a \rangle \to H$ is a homomorphism.
If you restrict the domain of $\phi$ to $\langle a \rangle = \{ a^k \;|\; k \in \mathbb{Z}\}$ (the cyclic subgroup generated by $a$), $\phi$ becomes a homomorphism.
This means that the order of $\phi(a)$ must divide the order of $a$.
Such maps are capturing some information about the cyclic subgroups of $G$. But I think that's about it.
Take the dihedral group $D_4 = \langle x,y \;|\; x^4=y^2=xyxy=1 \rangle = \{ 1,x,x^2,x^3,y,xy,x^2y,x^3y \}$ (i.e., $x$ is a rotation of 90 degrees and $y$ is a reflection).
Then you could map $\phi:D_4 \to D_4$ where $\phi(z)=z^2$. Then it is the case that $\phi(z^k)=z^{2k}=\phi(z)^k$ so $\phi$ preserves exponents.
But ``$\mathrm{Kernel}(\phi)$'' (stuff that maps to the identity) is $\{1,x^2,y,xy,x^2y,x^3y\}$ (everything but $x$ and $x^3$). This is not a subgroup. You no longer have that the fibers of $\phi$ are cosets and much of what you'd want to use a homomorphism to study no longer works.