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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?

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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.

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    $\begingroup$ Danny Ruberman's answer to mathoverflow.net/questions/344196/… provides an explicit example of using the two-variable Alexander polynomial to provide a counterexample to the original question. $\endgroup$ – dvitek Dec 13 '20 at 23:42
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    $\begingroup$ Thank dvitek, indeed the questions was already answered one year ago. However I like Marco's answer. Thanks to both of you. $\endgroup$ – Pierre Dehornoy Dec 14 '20 at 22:26

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