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?)
 A: To clarify, $\lambda=\operatorname{E}(X)=\operatorname{Var}(X)$.  The normal approximation is $N(\mu=\lambda,\sigma^2=\lambda)$ when $\lambda$ is "large".
$\lambda$ is often interpreted as an event arrival rate.
If you construct pseudorandom arrival times for the rate $\lambda_1 = \frac{1}{\text{1 hour}}$; you could "reuse" the constructed arrival times for a slower arrival process $\lambda_{24} = \frac{1}{\text{24 hour}}$ by scaling the pseudorandom intervals.
Defining $T_0$ as the interval start time, your arrival times for the faster process would stretch:  $$t_{i,24} = 24 ~(t_{i,1}-T_0)\quad .$$
In general,
$$t_{i,\lambda_2} = (t_{i,\lambda_1} - T_0)~ \frac{\lambda_1}{\lambda_2}\quad.$$
A: 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).
A: If you have a lot of independent Poisson values when $\lambda=1$ then is possible to construct independent Poisson values when $\lambda^\prime=n$ for any positive integer $n$ by adding $n$ of your original values together
To take Stephan Kolassa's example with $n=5$, by simulating $500000$ cases of $\lambda=1$ to generate $100000$ cases of $\lambda^\prime=5$, and then comparing to the actual distribution in red, you could use this R code:
set.seed(2021)
n <- 5 
samplesize <- 10^5
x1 <- rpois(n * samplesize, lambda=1)
xn <- rowSums(matrix(x1, ncol=n))
table(xn)
# xn
#    0     1     2     3     4     5     6     7     8     9    10    11    12 
#  693  3403  8359 14001 17540 17385 14758 10594  6452  3642  1724   869   367 
#   13    14    15    16    17    18 
#  134    54    14     9     1     1 
plot(table(xn) / samplesize)
m <- min(xn):max(xn)
points(m, dpois(m, lambda=n), col="red")


which is pretty close for a simulation
