Find a transformation such that the Fisher Information is constant in terms of the parameter. Say you have a Gamma distribution with parameters $\alpha$ known and $\theta$ unknown. Find a transformation of $\theta$, $\eta=g(\theta)$ such that ${\cal I} (\eta)$, the Fisher Information, is constant in terms of $\eta$.
My problem is that the only way I can do this is by trial and error. At the moment I haven't been able to come up with any transformations that make this work. Is there a method that this can be done?
 A: What this is asking you for is the normalizing transform, since the fisher information is the second derivative of the log likelihood, the transformation must result in a quadratic log-likelihood in $\eta$.
Take a look at Section 8.4 of this link, especially p.270. Has what you are looking for. I will use that formula in the below:
$\mathcal{I}(\eta)=\frac{\mathcal{I}(\theta)}{(\partial\eta/\partial\theta)^2}$ [From Pawitan 2001, p.270]
Re-state goal mathematically:
$\frac{\partial\mathcal{I}(\eta)}{\partial\eta}=0=\frac{\partial}{\partial\eta}[\frac{\mathcal{I}(\theta)}{(\partial\eta/\partial\theta)^2}]=\frac{\partial}{\partial\eta}[\mathcal{I}(\theta){(\partial\theta/\partial\eta)^2}]=2\mathcal{I}(\theta)(\partial\theta/\partial\eta)\frac{\partial^2\theta}{\partial\eta^2}+(\partial\theta/\partial\eta)^2\frac{\partial\mathcal{I}(\theta)}{\partial\theta}\frac{\partial\theta}{\partial\eta}$
Divide out the common $\partial\theta/\partial\eta$ to get the nonlinear differential equation:
$2\mathcal{I}(\theta)\frac{\partial^2\theta}{\partial\eta^2}+(\frac{\partial\theta}{\partial\eta})^2\frac{\partial\mathcal{I}(\theta)}{\partial\theta}=0$
Take reciprocals of each derivative to get a non-linear PDE in terms of $\eta(\theta)$:
$-2\mathcal{I}(\theta)(\frac{\partial^2\eta}{\partial\theta^2})(\frac{\partial \eta}{\partial \theta})^{-3}+(\frac{\partial\eta}{\partial\theta})^{-2}\frac{\partial\mathcal{I}(\theta)}{\partial\theta}=0$ 
Simplyfing we get: 
$\frac{\partial\mathcal{I}(\theta)}{\partial\theta}(2\mathcal{I}(\theta))^{-1}=(\frac{\partial^2\eta}{\partial\theta^2})(\frac{\partial \eta}{\partial \theta})^{-1}$ 
You'll need to solve this PDE for your  specific problem to get the answer. Not a trivial task.
A: In fact, it is a trivial task ! Denoting by prime the derivative with respect to $\theta$, the differential equation writes
\begin{equation}
\frac{\mathcal{I}^\prime(\theta)}{2 \mathcal{I}(\theta)} = \frac{\eta^{\prime\prime}(\theta)}{\eta^{\prime}(\theta)}.
\end{equation}
Integration is straightforward
\begin{equation}
\log(\eta^{\prime}(\theta))= \frac{1}{2}\log(\mathcal{I}(\theta)) + \text{cst}
\end{equation}
which leads to
\begin{equation}
\eta(\theta) \propto \int_0^\theta \sqrt{\mathcal{I}(t)} \mathrm{d}t + \text{cst}.
\end{equation}
