Transform "Standard Poisson" to any Poisson If I use a RNG to generate a set of standard normal draws $(Z_i)$, anyone can get samples from a normal distribution with $(\mu, \sigma^2)$ of their choosing via $X_i=Z_i.\sigma+\mu$. This doesn't require them to do any random number generating themselves (i.e. they don't need a RNG)
Is the same thing possible for Poisson distributions? i.e. If I generate a set of Poisson draws with $\lambda=1$ (or some other $\lambda$), is it possible for another person to get Poisson samples with whatever $\lambda$ they want, without having to use an RNG?
(I realise you could use the normal approximation $N(\lambda,\sqrt\lambda)$ if $\lambda$ is large, but what if this isn't the case?)

No, that is not possible.
For instance, assume we want to "transform" Poisson realizations with $\lambda=1$ to Poisson samples with $\lambda'=5$. The PMF at $0$ for $\lambda=1$ is $\frac{1}{e}\approx 0.368$, so about 36.8% of the original samples will be $0$. But the cumulative distribution function for $\lambda'=5$ is only $0.265$ for $x=3$. That is, we would need somehow map an original observation of $0$ to transformed observations $0,1,2,3,4$ - and this in a way that satisfies the PMF for the new $\lambda'$. This is simply not possible without a RNG.
The same holds for "transformations" between any two discrete distributions (except of course for trivial cases).
