Isotopy classes on the disk and mapping tori Is the following true?
"The conjugacy classes of two homeomorphisms of the n-times punctured disk have isotopic representatives iff the associated mapping tori are homeomorphic."
By conjugacy class of a homeomorphism is meant the homeomorphism up to (orientation preserving) topological changes of coordinates. The isotopies are not required to keep the boundary pointwise fixed. Thus the question may be rephrased as:
Is it true that, up to changes of coordinates and isotopies, two homeomorphisms of the punctured disk are the same iff their mapping tori are homeomorphic? The "only if" is true; it is the "if" that puzzles me.  
 A: As I said in the comment above, knots in $D^2\times S^1$ which have a Dehn filling giving $D^2\times S^1$ were classified by Gabai and Berge. Gabai proved that knots in $D^2\times S^1$ giving back $D^2\times S^1$ are either cables or 1-bridge braids (note that the original statement was given before the knot complement problem, so Gabai states that the knot could also lie in a ball, but now this is known to not occur). I didn't check, but I believe the cable case will not work: even though there are many solid torus Dehn fillings, the link complement has an infinite image of the mapping class group in the mapping class group of the boundary, so I believe these all give equivalent cablings (although I didn't check this). 
For 1-bridge braids in a solid torus with solid torus surgery, Gabai gave a partial classification, and Berge gave a complete classification. Gabai shows that the other 1-bridge braid will have the same winding number (Corollary 3.3), so the same number of strands. For some of these examples, Dehn filling gives back a 1-bridge braid in the solid torus of the same type. But Berge shows that most examples will give back a different braid. 
To be explicit, I took the braid from Figure 8 of Gabai's paper, which is a braid on 10 strands. I input this into SnapPy, which shows that the braid is hyperbolic, and $(63,1)$ Dehn filling on cusp $0$ gives a manifold with fundamental group $\mathbb{Z}$, which therefore must be $D^2\times S^1$. Moreover, the symmetry group is $\mathbb{Z}/2^2$, and there is a $\mathbb{Z}/2$ subgroup which preserves the cusps, and acts as an elliptic involution, therefore preserving slopes. So the two 1-bridge braids must be inequivalent, but have the same complement.

