I came up with the following proof, which I would like help checking and/or cleaning up:
Theorem: Let $X$ and $Y$ be $G$-sets, and let $\varphi:X\to Y$ be a $G$-equivariant map, i.e. $\varphi(gx) = g\varphi(x)$ for all $x\in X, g\in G.$ If the action is transitive on $Y$ then $|Y|$ divides $|X|.$
My attempt: $Y$ is isomorphic as a $G$-set to $G/H$ with the left multiplication action for some $H.$ Similarly the action on $X$ is transitive on each of the orbits $X_1,\dots, X_n,$ so each $X_i$ is also isomorphic to a $G$-set of the form $G/G_i$ with a left multiplication action. We are asked to show that
\begin{align*} |X|= |Y|\cdot k. \\ |X_1\sqcup\dots\sqcup X_n| = \frac{|G|}{|H|}\cdot k \\ |G|\left(\frac{1}{|G_1|} +\dots + \frac{1}{|G_n|}\right) = \frac{|G|}{|H|}\cdot k \\ \frac{|H|}{|G_1|} + \dots + \frac{|H|}{|G_n|} = k\\ \end{align*} so if we can show that $\frac{|H|}{|G_i|}$ is an integer we are done.
We pay attention to the two $G$-sets, $G/G_i$ and $G/H$ and the restriction of our $G$-equivariant map must clearly be $\tilde{\varphi}: G/G_i\to G/H, gG_i\mapsto gH.$ Now let $a\in G_i.$ I claim that $a\in H$ as well. Indeed $$aH=a\tilde{\varphi}(G_i)=\tilde{\varphi}(aG_i) = \tilde{\varphi}(G_i) = H$$ So $G_i$ is a subgroup of $H$ and the claim follows.
Is this proof correct? If so, can it be improved? Are there steps that should be worded differently, made shorter or more elegant? Are there easier ways of arriving at this result?
I think your proof is correct, and can only be improved by getting rid of the implicit finiteness hypothesis (of $X$ and $G$).
$X=\sqcup_{i\in I}X_i,$ $G_i\subset H,$ hence $|X_i|=|G/G_i|=|G/H||H/G_i|=|Y||H/G_i|$ and $$|X|=|Y|\sum_{i\in I}|H/G_i|,$$ where $I$ and each cardinal in this formula is allowed to be infinite.