Orbits of a group as finite copies of the orbits of its subgroup Let us consider an abelian group $G = A \oplus H$, where $A$ is a finite group. Let $G$ and $H$ act on a set $X$ and let us consider the $G$-orbits and $H$-orbits of $X$ respectively. 
Is it true that for every $G$-orbit, there are $|A|$ many isomorphic copies of $H$-orbit? Does $A$ have to be cyclic?
(The cardinalities of X and G both need not be finite.)
 A: Since you care about Abelian groups, the following proposition should be helpful, noting that for Abelian groups, transitive and faithful implies transitive and free (proof further down).
Proposition: Let $G$ act (wlog) transitively and freely on $X$ (on the left), let $H$ be a subgroup of $G$, then $X$ is a disjoint union of $[G:H]$ $H$-orbits, each of which is isomorphic as an $H$-set.
Proof: Since the orbits decompose $X$ into disjoint pieces for any group acting on it, $X$ is certainly a disjoint union of $H$-orbits. All we need to prove is that there are $[G:H]$ $H$-orbits, and that they are all isomorphic. 
Fix $x\in X$, then since $G$ acts transitively and freely on $X$, $X$ is isomorphic to $G$ as a left $G$-set by the map $G\to X$, $g\mapsto gx$. (The map is surjective by transitivity, and injective by freeness). Then under this isomorphism, $H$-orbits are just $Hg$ for $g\in G$. I.e. under the isomorphism, the $H$-orbits are the right cosets of $H$. Then the isomorphism from $Hg_1$ to $Hg_2$ is just right multiplication by $g_1^{-1} g_2$, which is certainly left $H$-linear.
This proves our claim.
Comments: We can assume transitivity, since we only care about $G$-orbits, so we can restrict our attention to a single $G$-orbit. We can assume faithfulness, because if $G$ does not act faithfully, we can quotient out by the subgroup, $N$, that acts trivially. Let $q$ be the quotient map, then we can apply the result to $q(H)\le G/N$. An example for why this assumption is necessary: consider your original question, now assume $A$ acts trivially, then $G$ and $H$ orbits are the same, so the statement you are asking about is false. Thus we need this assumption to say anything meaningful.
Comment on edit: As pointed out in the comments, the prior statement of the proposition was false. Previously I had confused the statement of the proposition and had faithful instead of free, but faithful is too weak of an assumption in general, so I've fixed it to what I'm fairly confident is now correct.
Lemma: When $G$ is abelian, if $G$ acts transitively and faithfully on $X$, then it acts freely on $X$.
Proof: Suppose $gx=x$, then by transitivity for all $y\in X$, there is some $h_y\in G$ such that $h_yx=y$. Then $gy=gh_yx=h_ygx=h_yx=y$ for all $y\in G$, but $G$ acts faithfully, so $g$ must be the identity element. Thus if $gx=x$, $g$ is the identity. This is what is means for $G$ to act freely.
Edit in response to update that was posted as an answer:
First of all, I assume you mean $\oplus$ rather than $\otimes$. Then in the case you've given, the situation is the following: We have $R$ a ring, such that $H\subseteq G\subseteq R^\times$. Even if $R^\times$ and $G$ are not commutative, if $gu=u$ for $g\in G$, $u\in U$, we can cancel $u$ to get $g=1$. Thus $G$ acts freely on $R^\times$, so when we restrict to $G$-orbits, $G$ acts transitively and freely on $G$-orbits. Thus we can directly apply the proposition to get that $G$-orbits can be decomposed into $[G:H]$ $H$-orbits each isomorphic as $H$-sets.
A: Here is a very specific example: Let us consider a multiplicative group $G = U_n \times H$, where $U_n$ is the cyclic group containing $n^{th}$ roots of unity. Let us consider a ring $R$ containing $G$ as a set. Let $G$ act on $R$ by multiplication.
We have, for every $g \in G$, $g = \omega h$ where $\omega \in U_n$ and $h \in H$. So, for each $x \in R$, the $G$-orbit of $x$ corresponds to the following $H$-orbits:
$xH$, $x\omega H$, ... , $x\omega^{n-1}H$
and each of these $H$-orbits are isomorphic under the multiplication map $x \mapsto x \omega$.
