Let $S$ be a countable dense subset of $\mathbb R$. Must there exist a homeomorphism $f: \mathbb R \rightarrow \mathbb R$ such that $f(S) = \mathbb Q$? More weakly, must $S$ be homeomorphic to $\mathbb Q$?
Two countable totally ordered, densely ordered sets without endpoints are isomorphic—this is a theorem of Cantor (Gesammelte Ahbandlungen, chp. 9, page 303 ff. Springer, 1932) Thus two countable dense subsets of $\mathbb R$ are homeomorphic, since their topology is induced by their orders.
Now, if $A$, $B\subset\mathbb R$ are countable dense subsets, fix an order isomrphism $f:A\to B$ and extend it by continuity. What you get is an homeomorphism $\mathbb R\to\mathbb R$.