Counterexamples are easily constructed using the Thurston norm. In fact, any example of a fibered, oriented, closed 3-manifold $M$, with a fiber of genus $\ge 2$, and with 2nd homology of rank $\ge 2$, gives counterexamples.

The Thurston norm on $H_2(M;\mathbb{R})$ has a polyhedral unit ball, and there is a symmetric set of top dimensional faces, called "fibered faces", such that a homology class contained in the integer lattice $H_2(M;\mathbb{Z})$ of $H_2(M;\mathbb{R})$ is represented by a fiber of a fibration over $S^1$ if and only if that class is in the interior of the cone of a fibered face. Furthermore, inside the cone on a fibered face, the value of the norm is given by a rationally defined linear functional $x : H_2(M;\mathbb{R}) \to \mathbb{R}$ such that if $S$ is in the integer point in that cone then $x(S)=-\chi(S)$.

Suppose $M$ fibers, and suppose $\text{rank}(M) \ge 2$. Then there exists a fibered face, with corresponding linear functional $x$. Consider nonempty integer level sets of the form $L_k = x^{-1}(k) \cap H_2(M;\mathbb{Z})$ ($k$ must be even for $L_k$ to be nonempty). As $k \to +\infty$, clearly the cardinality of $L_k$ gets larger and larger. Thus one obtains an arbitrarily large number of fibers all of the same Euler characteristic, but representing pairwise distinct homology classes.

But the group of homeomorphisms modulo isotopy of $M$ is finite, say of cardinality $A$ and so at most $A$ homology classes can represent conjugate mapping classes.

Counterexamples are easily constructed using the Thurston norm. In fact, any example of a fibered, oriented, closed 3-manifold $M$, with a fiber of genus $\ge 2$ and with pseudo-Anosov monodromy, and with 2nd homology of rank $\ge 2$, gives counterexamples.

The Thurston norm on $H_2(M;\mathbb{R})$ has a polyhedral unit ball, and there is a symmetric set of top dimensional faces, called "fibered faces", such that a homology class contained in the integer lattice $H_2(M;\mathbb{Z})$ of $H_2(M;\mathbb{R})$ is represented by a fiber of a fibration over $S^1$ if and only if that class is in the interior of the cone of a fibered face. Furthermore, inside the cone on a fibered face, the value of the norm is given by a rationally defined linear functional $x : H_2(M;\mathbb{R}) \to \mathbb{R}$ such that if $S$ is in the integer point in that cone then $x(S)=-\chi(S)$.

We are supposing that $M$ fibers and that $\text{rank}(M) \ge 2$, and so there exists a fibered face, with corresponding linear functional $x$. Consider nonempty integer level sets of the form $L_k = x^{-1}(k) \cap H_2(M;\mathbb{Z})$ ($k$ must be even for $L_k$ to be nonempty). As $k \to +\infty$, clearly the cardinality of $L_k$ gets larger and larger. Thus one obtains an arbitrarily large number of fibers all of the same Euler characteristic, but representing pairwise distinct homology classes.

But $M$ is a closed hyperbolic 3-manifold and hence its group of homeomorphisms modulo isotopy of $M$ is finite, say of cardinality $A$. So at most $A$ homology classes can represent conjugate mapping classes.

Counterexamples are easily constructed using the Thurston norm. In fact, any example of a fibered $M$ with first homology of rank $\ge 2$ gives counterexamples.

The Thurston norm on $H_1(M;\mathbb{R})$ has a polyhedral unit ball, and there is a symmetric set of top dimensional faces, called "fibered faces", such that a homology class contained in the integer lattice $H_1(M;\mathbb{Z})$ of $H_1(M;\mathbb{R})$ is represented by a fiber of a fibration over $S^1$ if and only if that class is in the interior of the cone of a fibered face. Furthermore, inside the cone on a fibered face, the value of the norm is given by a rationally defined linear functional $x : H_1(M;\mathbb{R}) \to \mathbb{R}$ such that if $S$ is in the integer point in that cone then $x(S)=-\chi(S)$.

Suppose $M$ fibers, and suppose $\text{rank}(M) \ge 2$. Then there exists a fibered face, with corresponding linear functional $x$. Consider nonempty integer level sets of the form $L_k = x^{-1}(k) \cap H_1(M;\mathbb{Z})$ ($k$ must be even for $L_k$ to be nonempty). As $k \to +\infty$, clearly the cardinality of $L_k$ gets larger and larger. Thus one obtains an arbitrarily large number of fibers all of the same Euler characteristic, but representing pairwise distinct homology classes.

But the group of homeomorphisms modulo isotopy of $M$ is finite, say of cardinality $A$ and so at most $A$ homology classes can represent conjugate mapping classes.

Counterexamples are easily constructed using the Thurston norm. In fact, any example of a fibered, oriented, closed 3-manifold $M$, with a fiber of genus $\ge 2$, and with 2nd homology of rank $\ge 2$, gives counterexamples.

The Thurston norm on $H_2(M;\mathbb{R})$ has a polyhedral unit ball, and there is a symmetric set of top dimensional faces, called "fibered faces", such that a homology class contained in the integer lattice $H_2(M;\mathbb{Z})$ of $H_2(M;\mathbb{R})$ is represented by a fiber of a fibration over $S^1$ if and only if that class is in the interior of the cone of a fibered face. Furthermore, inside the cone on a fibered face, the value of the norm is given by a rationally defined linear functional $x : H_2(M;\mathbb{R}) \to \mathbb{R}$ such that if $S$ is in the integer point in that cone then $x(S)=-\chi(S)$.

Suppose $M$ fibers, and suppose $\text{rank}(M) \ge 2$. Then there exists a fibered face, with corresponding linear functional $x$. Consider nonempty integer level sets of the form $L_k = x^{-1}(k) \cap H_2(M;\mathbb{Z})$ ($k$ must be even for $L_k$ to be nonempty). As $k \to +\infty$, clearly the cardinality of $L_k$ gets larger and larger. Thus one obtains an arbitrarily large number of fibers all of the same Euler characteristic, but representing pairwise distinct homology classes.

But the group of homeomorphisms modulo isotopy of $M$ is finite, say of cardinality $A$ and so at most $A$ homology classes can represent conjugate mapping classes.