Looking at Automorphisms of Subgroups of $(\mathbb R,+)$ With Positive Slope Let $(R,+)$ be a non-trivial subgroup of $(\mathbb R,+)$.
We say that an automorphism $\phi$ of $R$ has positive slope if
$\tag 1 \phi(R \cap \mathbb R^{\gt 0}) \subset \mathbb R^{\gt 0}$

What conditions must be put on $R$ to guarantee that
For every $x,  y \in R \cap \mathbb R^{\gt 0}$ there is one and only one automorphism of positive slope $\phi$
such that $\phi(x) = y$.

My Work
I know that when $R = \mathbb R$ the automorphism can always be found. I suspect that it is true provided $R$ can be n̶a̶t̶u̶r̶a̶l̶l̶y̶ endowed with a multiplicative structure using the automorphisms. You can call any positive element in $R$ the unit of measure $1$ and you get a field (the automorphisms form a group).
Of interest here is the wikipedia article on Cauchy's functional equation.
 A: Note that we can safely assume $x<y$; otherwise, just look at the inverse map.
First, note that $R$ must be dense in order to have any nontrivial automorphisms of positive slope. (Put another way, the group $\mathbb{Z}$ has exactly two automorphisms, only the identity has positive slope, and any discrete subgroup of $R$ is isomorphic - indeed, ambiently isomorphic - to $\mathbb{Z}$.)
From now on, $R$ is a dense subgroup of $\mathbb{R}$.
Second, note that any automorphism of positive slope must be order-preserving on positive elements: if $x>y>0$ but $f(x)<f(y)$, then we would have $x-y>0$ but $f(x-y)<0$.
This means that any automorphism $f$ of $R$ with positive slope extends uniquely to an automorphism $\hat{f}$ of $\mathbb{R}$ with positive slope. Namely, given $r\in \mathbb{R}_{>0}$ we pick an increasing sequence $a_1<a_2<...$ of elements of $R$ with $\lim_{i\rightarrow\infty}a_i=r$. Then $f(a_1)<f(a_2)<...$, and this sequence is bounded above (since $R_{>r}\not=\emptyset$), so it has some limit $s$. It's now easy to check that two different sequences of $a$s yield the same $s$ and that the map $r\mapsto s$ is an automorphism of $\mathbb{R}$ with positive slope.
But the automorphisms of $\mathbb{R}$ of positive slope are exactly the maps of the form $x\mapsto cx$ for $c\in\mathbb{R}_{>0}$, and in particular are determined by where they move any single nonzero value; so if an automorphism of $R$ with positive slope moving $x$ to $y$ exists, it must be unique.

So now we're left only with the existence task. But in light of the classification above, this is easy: $R$ has the desired property iff for every $0<x<y$ in $R$ the map $z\mapsto {yz\over x}$ is defined on $R$ - that is, iff for any $x,y,z$ in $R$ with $0<x<y$ we have ${yz\over x}\in R$. 
Trivially any subfield of $\mathbb{R}$ has this property; conversely, if $1\in R$ and $R$ has this property then $R$ is a field. Any such $R$ can be "multiplicatively scaled" to an $\hat{R}$ containing $1$; when we do this, we preserve the above property and so get a field. Conversely, by the same reasoning any multiplicative scaling of a field has this property (e.g. $R=\{q\pi:q\in\mathbb{Q}\}$).
So this gives a complete answer to your question:

$R$ has the property in question iff $R=\{cx: x\in F\}$ for some field $F\subseteq\mathbb{R}$ and some nonzero constant $c\in\mathbb{R}$.

