Are there any implications between "pair-homogeneity" and "scale homogeneity" for modules over an arbitrary commutative ring? 
Definition. Suppose $R$ is a ring and $X$ is a module over $R$. Then:
  
  
*
  
*$X$ is scale-homogeneous iff for all non-zero $a \in R$, the mapping $X \rightarrow X$ given by $x \mapsto ax$ is an automorphism. (Note: this doesn't imply that $a$ is a unit, because we don't require that the inverse of the function $x \mapsto ax$ be of the form $x \mapsto bx$ for some $b \in R$.)
  
*$X$ is pair-homogeneous iff for all non-zero $x,y \in X$, there is an automorphism $\varphi$ of $X$ such that $y=\varphi x.$

We have:

Proposition. Let $R$ denote a commutative ring. Then TFAE.
  
  
*
  
*$R$ is a field.
  
*$R$ is scale-homogeneous as a module over itself.
  
*$R$ is pair-homogeneous as a module over itself.
  

What I'd like to know is how these conditions relate for an arbitrary module over an arbitrary commutative ring.

Question. Suppose $R$ is a commutative ring and $X$ is a module over $R$. Are there any implications between the statements: "$X$ is scale-homogeneous" and "$X$ is pair-homogeneous"?

 A: Proposition: If $M$ is nonzero, $M$ is scale-homogeneous iff $R$ is a domain and $M$ is a torsion-free, divisible module. 
Proof: Obviously the condition that multiplication by nonzero ring elements is injective implies that no element of $M$ has a nonzero annihilator. It also implies that the composition of scaling is injective, so $ab\neq 0$ whenever $a,b$ are nonzero. The multiplication map being onto $M$ indicates $M$ is divisible.
Proposition: $M$ is pair homogeneous iff it is a simple module over its endomorphism ring. If $M_R$ is faithful, then $R$ embeds in $E=End(M_R)$. Now $_EM$ is already a faithful module, so $E$ embeds into $End(_EM)^{op}$ which is a division ring. The net result is an embedding of $R$ into a division ring, so $R$ would be a domain in that case. If $M_R$ isn't faithful, only a quotient of it embeds in the division ring and the annihilator (the kernel of the embedding) is a prime ideal in $R$.)
A: Theorem. Let $R$ be a commutative ring. If an $R$-module $X$ is scale-homogeneous, then $R$ is pair-homogeneous.
Proof. Let $X$ be a scale-homogeneous $R$-module. We need to show that $X$ is pair-homogeneous.
First, assume that $R$ is not an integral domain. Then, there are two nonzero elements $a$ and $b$ of $R$ such that $ab = 0$. Consider these $a$ and $b$. Both $a$ and $b$ act on $X$ as automorphisms (since $X$ is scale-homogeneous). Therefore, $ab$ acts on $X$ as an automorphism. Since $ab = 0$, this shows that the zero map $X \to X$ is an automorphism of $X$. Thus, $X = 0$, so that $X$ is pair-homogeneous.
Now, forget that we assumed that $R$ is not an integral domain. We thus have proven the Theorem in the case when $R$ is not an integral domain. We now WLOG assume that $R$ is an integral domain. Let $S = R \setminus \left\{0\right\}$. Then, $S$ is a multiplicative subset of $R$ (since $R$ is an integral domain), and thus the localization $R_S$ is well-defined.
The action of $R$ on the $R$-module $X$ can be written as a ring homomorphism $R \to \operatorname{End}X$ (where $\operatorname{End}$ means $\operatorname{End}_{\mathbb{Z}}$). This homomorphism sends all elements of $S$ to automorphisms of $X$ (since $X$ is scale-homogeneous), i.e., to invertible elements of $\operatorname{End}X$. Hence, it factors through the localization $R_S$ (by the universal property of this localization). In other words, the $R$-module $X$ is the restriction of an $R_S$-module, which we also call $X$ (since it has the same ground set as $X$).
But $R_S$ is the localization of the integral domain $R$ at the set $S$ of all its nonzero elements. Hence, $R_S$ is a field (in fact, you can check that any nonzero element $a/b$ of $R_S$ must have $a\neq 0$, and therefore has $b/a$ as its inverse). Thus, the $R_S$-module $X$ becomes an $R_S$-vector space.
Now, let $x, y \in X$ be nonzero. In order to prove that $X$ is pair-homogeneous, we must show that there exists an automorphism $\varphi$ of the $R$-module $X$ sending $x$ to $y$. But we can extend the one-element family $\left(x\right)$ to a basis $\left(x_i\right)_{i \in I}$ of the $R_S$-vector space $X$, and we can extend the one-element family $\left(y\right)$ to a basis $\left(y_i\right)_{i \in J}$ of the $R_S$-vector space $X$. Since all bases of a vector space have the same cardinality, we have $\left|I\right| = \left|J\right|$, and thus we can WLOG assume that $I=J$. Assume this. Then, we can clearly find an automorphism $\varphi$ of the $R_S$-vector space $X$ which sends each $x_i$ to the corresponding $y_i$. This $\varphi$ is an automorphism of the $R$-module $X$, and sends $x$ to $y$; thus we are done. The Theorem is proven.
