I am trying to solve:
Let $G,H$ be cyclic groups, generated by elements $x,y$. Determine the condition on the orders $m,n$ of $x$ and $y$ so that the map sending $x^i \rightsquigarrow y^i$ is a group homomorphism.
Here is my attempt.
We require that the map $\varphi: \underset{x^i}{G} \underset{\longmapsto}{\to} \underset{y^j}{H}$ is well-defined. That is, if $x^i = x^j$ for $i,j \in \mathbb{Z}$, then $y^i = y^j$. That is, if $i \equiv j \text{ (mod $m$)}$, then $i \equiv j \text{ (mod $n$)}$, i.e., if $i - j = \alpha m$ for some $\alpha \in \mathbb{Z}$, then $i - j = \beta n$ for some $\beta \in \mathbb{Z}$. As $i,j$ are arbitrary and any integer can be written in the form $i - j$, this means that for all $z \in \mathbb{Z}$, $m \mid z$ implies that $n \mid z$, which is true provided that $n \mid m$. To show that this condition is sufficient, let $m = nk$ for some $k \in \mathbb{Z}$ and suppose that $x^i = x^j$ for some $i,j \in \mathbb{Z}$. Then we have \begin{align*} i - j = \gamma m = \gamma (nk) = n(\gamma k) \end{align*} for some $m \in \mathbb{Z}$, so $i \equiv j \text{ (mod $n$)}$, so $y^i = y^j$, hence $f(x^i) = f(x^j)$, so $\varphi$ is well-defined. I claim that $\varphi$ is then a homomorphism. Indeed, for any $i,j \in \mathbb{Z}$, we have \begin{align*} \varphi\left(x^i x^j\right) = \varphi\left(x^{i+j}\right) = y^{i+j} = y^i y^j = \varphi\left(x^i\right) \varphi\left(x^j\right), \end{align*} as desired.
Is the logic correct? The only thing I am still not completely certain about is the pairing of $i-j$ with some arbitrary integer $z$, but there is surely at least a surjective mapping from differences of integers with the integers themselves, so I believe the logic is valid.
I believe you have the right condition there: $n\mid m $.
That's because the order of $\varphi (x) =y $ is $n $; but it must also divide $m $, since $(\varphi (x))^m=\varphi (x^m)=\varphi (e)=e $.
On the other hand, well-definedness doesn't seem to be much of an issue (anymore) because the homomorphism property will ensure it.
So, the condition is necessary and sufficient.