There is an example of an Akbulut cork, proved to be a strong cork by Dai-Hedden-Mallick, $W= W(0,1)$ whose boundary $M$ is $+1$ surgery on the Stevedore knot, and has a mapping class group of order 4 generated by two involutions $S$ and $T$. If one doubles $W$ along $M$, then one gets $S^4=-W\cup W$, and hence an embedding of $M$ into $S^4$, since $W$ is a Mazur manifold (described first by Akbulut-Kirby).
(Proposition 1 of Mazur proves that any such manifold has double diffeomorphic to $S^4$)

Akbulut also shows that $-W \cup_S W$ obtained by gluing two copies by $S$ is diffeomorphic to $S^4$, giving another embedding of $M$ into $S^4$, and same for $-W\cup_T W$. Suppose the two embeddings are isotopic, in particular there is a diffeomorphism taking one embedding to the other. Then this diffeomorphism takes the $W$ on each side to each other. We may think of one $W$ as fixed, and the diffeomorphism extends over the other $W$. But the other two copies of $W$ are glued by the identity and $S$ respectively, so we see that the involution $S$ extends to a diffeomorphism of $W$, a contradiction. Thus we have two copies of $M$ embedded in $S^4$ which are not isotopic, and which have diffeomorphic complements.

One comment: if the isotopy takes the $W$ on one side to the other copy of $W$ in such a way that the diffeomorphism is not isotopic to the identity on $M$, then it must be the involution $ST$ in the mapping class group of $M$, since $S$ and $T$ don’t extend. Then the two mapping classes will differ by $T$, which still doesn’t extend over the other copy of $W$.

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