Yes. If two pseudo-Anosov mapping classes are conjugate then they must have the same dilatation. So take any pseudo-Anosov $f$ and any mapping class $h$ not in the centraliser of $f$ and let $g = h f h^{-1}$. Then $f$ and $g$ are distinct, both pseudo-Anosov and have the same dilatation.

However, even if you require that $f$ and $g$ are not conjugate the answer is still yes. For example on the twice-punctured torus, let $f = T_aT_bT_c^{-1}$ and $g = T_aT_b^{-1}T_c^{-1}$ where these are Dehn twists about the curves $a$, $b$, $c$ shown below.

These both have dilatation the largest real root of

$$ 1 - 2x -2x^3 + x^4, $$

which is $\approx 2.2966$, but are not conjugate. You can check this by using flipper and the Python code:

```
>>> import flipper
>>> S = flipper.load('S_1_2')
>>> f = S.mapping_class('abC')
>>> g = S.mapping_class('aBC')
>>> f.dilatation() == g.dilatation()
True
>>> f.is_conjugate_to(g)
False
```

## Generating:

The following code was used to find this example. It lists all the different mapping classes with word length at most three. It filters out the pseudo-Anosov ones and sorts them by dilatation. We group these by dilatation and then look for things in a group that are not conjugate to the first one.

```
>>> X = set(S.all_mapping_classes(length=3, letters=list('abcABC')))
>>> Y = sorted([x for x in X if x.is_pseudo_anosov()], key=lambda x: x.dilatation())
>>> Z = [list(b) for _, b in groupby(Y, key=lambda y: y.dilatation())]
>>> Z
[[a.b.C, a.C.b, a.C.B, a.B.C], [a.c.B, a.B.c]]
>>> for P in Z: print([p.is_conjugate_to(P[0]) for p in P])
[True, True, False, False]
[True, True]
```

It's not too hard to see why this example works. Namely $f$ is actually conjugate to $g^{-1}$ so they must have the same dilatation. But (the stable lamination of $f$) is lacking the symmetry needed for $f$ to be conjugate to $f^{-1}$, so $f \not \equiv f^{-1} \equiv g$:

```
>>> f.is_conjugate_to(g**-1)
True
```