TITLE: Homeomorphic but Non-Conjugate Mapping Tori
QUESTION [10 upvotes]: Suppose we fix a genus $g$ closed surface $S$. Let $f, g \in Map(S)$ be conjugate, for $Map(S)$ the mapping class group of $S$. Then I know that $M_f$ (the mapping torus of $M$ with monodromy $f$) is homeomorphic to $M_g$. 
Now, let $M$ is a mapping torus with monodromy $f$, and $N$ is a mapping torus with monodromy $g$. Suppose we know that $M_f$ is abstractly homeomorphic to $N_g$ (not homeomorphic as surface bundles over $S^1$, just homeomorphic as 3-manifolds). If $M_f$ and $N_g$ fiber over $S^1$ with homeomorphic fibers, then need $f, g$ be conjugate? 
I think the answer is "no", but I'm not sure why, and I'm not sure how to construct a counterexample. Any help would be greatly appreciated!

REPLY [3 votes]: To add an explicit counter-example, consider the curves $a$, $b$ and $c$ on the twice punctured torus shown below.

The mapping classes $f = T_a T_a T_b T_c^{-1}$ and $g = T_a T_a T_b^{-1} T_c^{-1}$ are not conjugate (they don't even have the same dilatation). However their mapping tori $M_f$ and $M_g$ are homeomorphic. You can check this by using the latest versions of flipper, SnapPy and the Python code:
>>> import flipper
>>> S = flipper.load('S_1_2')
>>> f = S.mapping_class('aabC')
>>> g = S.mapping_class('aaBC')
>>> f.is_conjugate_to(g)
False
>>> f.dilatation(), g.dilatation()
(2.618033?, 3.254263?)
>>> import snappy
>>> M_f = snappy.Manifold(f.bundle())
>>> M_g = snappy.Manifold(g.bundle())
>>> M_f.is_isometric_to(M_g)
True