I have (what I believe to be) a partial proof of the problem below. I do not; however, know how to treat the $\eta = 1, \lambda \neq 1$ case.
For any $\eta > 0$ let
\begin{equation} h_\eta(z) = - \eta \overline{z}. \end{equation}
Determine for which pairs $\eta, \lambda > 0$ the mappings $h_\eta, h_\lambda$ are conjugate in $\text{Mob}(\mathbb{H)}$.
Partial proof: Suppose that the mappings $h_\eta, h_\lambda$ are conjugate in $\text{Mob}(\mathbb{H})$. It follows that:
\begin{equation} \tau(h_\eta) = \tau(h_\lambda). \end{equation}
Hence (w.l.o.g.),
\begin{equation} \eta + \frac{1}{\eta} + 2 = \lambda + \frac{1}{\lambda} + 2. \end{equation}
Therefore, either $\eta = \lambda$, or $\eta = \frac{1}{\lambda}$.
For your reference:
$\tau(f)$ denotes the squared trace of $f \in \text{Mob}(\mathbb{C})$.