 Solution

False. Take the finite, hence countable, cyclic group $C_6=⟨a\mid a^6=1⟩$. Its elements $a^3$ and $a^2$ have orders two and three. A group embedding preserves the order of every element, whereas conjugate group elements have the same order: $(s^{-1}xs{)}^r=s^{-1}{x}^{r}s$. Therefore their images cannot be conjugate in any containing group. This gives the required obstruction. The torsion-free group hypothesis in Question 5 avoids this conflict of element orders.