This answer says that each p-adic number is a ratio of endomorphisms of a subgroup of a quotient of $\Bbb Q$.

If I wanted to realise this ratio of endomorphims for the $2$-adic number $-\dfrac13$ say, then given the context I guess the subgroup of a quotient of the rationals is $G=\Bbb Z[\frac12]/\Bbb Z$ since that's the Prufer 2-group.

Would two endomorphims $e_1,e_2$ on $G$ to yield $-\dfrac13$ simply be $e_1=-1\cdot G$ and $e_2=3\cdot G$, with conjugation to form the group operation?

There is an obvious case for the above, but it feels like the answer should be more complicated. The definition of an endomorphism as opposed to an automorphism or an isomorphism seems to require it's not necessarily reversible, therefore presumably it need not surject.