Suppose $L \subset S^3$ is a link consisting of two knots $J,K$ and $J,K$ have the same knot type. If $J,K$ are unknots, can we conclude the existence of a homeomorphism $\varphi :S^3\to S^3$ which interchanges $J,K$ ?
If $J,K$ are not trivial, the conclusion is not true. For example, we can modify the example 3F13 in Rolfsen's book to make $J,K$ asymmetric trefoils.
Assume $L$ is an oriented link, and let $[\mu_J],[\mu_K]\in H_1(S^3-L)$ represent meridians of components $J$ and $K$. A self-homeomorphism $\varphi$ of $S^3$ interchanging $J$ and $K$ on homology swaps $[\mu_J]$ and $[\mu_K]$, possibly multiplied by $-1$, corresponding to the orientation reversal of a component.
What this means for the multivariable Alexander polynomial $\Delta_L(t_J,t_K)$ is that $\Delta_L(t_J,t_K)=\Delta_{L'}(at_K,bt_J)$ where $a,b=\pm 1$ and $L'$ is $L$ with some components possibly reversed.
I paged through the LinkInfo database until I found a $2$-component link whose multivariable Alexander polynomial was invariant under component reversals and which did not have this symmetry. L7a1 was the first, and its components happened to be unknots.
$$\Delta_{\mathrm{L7a1}}(t_1,t_2)=1-t_1-2t_2+2t_1t_2+2t_2^2-2t_1t_2^2-t_2^3+t_1t_2^3$$