Given a 2-component link in $S^3$ whose components are trivial knots, is it always possible to find a homeomorphism of $S^3$ that exchanges the components?


I guess the answer is "no" (but I could not find a counterexample), so here is a second question:


Which link-invariant could prevent the existence of such an exchanging homeomorphism?



Exchanging the components of a two-component link
Indeed, as you suspect, the answer is no. For instance, take the link $L$ obtained from the Hopf link by doing a $(2,1)$-cable of one component and a Whitehead double of the other.


A way of telling them apart is to look at the JSJ decomposition of the complement: each piece contains one component, and one component is Seifert fibred, while the other is hyperbolic.


A reflection of this fact is that if you twist each of the components with respect to the other, you obtain different knots. (By "twist" here I essentially mean "do $\pm1$-surgery along one component.)


I also think that the 2-variable Alexander polynomial (or branched covers, coloured signatures...) should do the trick.


