If $Y_1, Y_2, ..., Y_n$ denote a random sample from a Poission distribution with mean $\lambda$, how can you determine a minimal sufficient statistic for $e^{-\lambda}$?
Using the Lehmann–Scheffé method I can find a minimal sufficient statistic for $\lambda$ but not $e^{-\lambda}$.
i.e, the ratio $\displaystyle{\frac{L(x_1,\ x_2, ...,\ x_n)}{L(y_1,\ y_2, ...,\ _yn)} = \frac{\lambda^{\sum_{i=1}^{n} x_i} \ e^{-n\lambda}}{\prod_{i=1}^{n} x_i!} . \frac{\prod_{i=1}^{n} y_i!}{\lambda^{\sum_{i=1}^{n} y_i}\ e^{-n\lambda}}}$
is free of $\lambda$ iff $\sum_{i=1}^{n} x_i$ = $\sum_{i=1}^{n} y_i$. So, $\sum_{i=1}^{n} y_i$ is a minimal sufficient statistic for $\lambda$. But the ratio is always free of $e^{-\lambda}$.
Note that $g(\lambda) = e^{-\lambda}$ is a one-to-one transformation of $\lambda$, thus if $\sum_{i=1}^n y_i$ sufficient for $\lambda$, then it is also sufficient for $g(\lambda)$.
Quick heuristic reasoning: If for any one to one (actually, for any measurable function, but let us restrict the discussion for one to one functions) function $g$ of $\lambda$, MLE of $g$ is the same function $g$ of the MLE of $\lambda$, and the MLE is always a function of minimal sufficient statistic, thus the same statistic is sufficient for $g(\lambda)$ as well.
For a formal proof you can use reparamtrization of the likelihood function by denoting $\theta = e^{-\lambda}$, and then to show that $\mathcal{I}_{\sum x_i}(\theta) = \mathcal{I}_{X_1,...,X_n}(\theta)$.