Is 'poisson-izing' a feature a useful method? When using linear LASSO regression, it is wise to normalize the features; for example, by subtracting the sample mean of the feature and dividing by the sample standard deviation: 
$$ X' = \frac{X - \bar{X}}{s}$$
where X is a column of features; $\bar{X}$ is the  mean of that column and $s$ is the standard deviation. Since the residuals are assumed to be Gaussian for linear least squares regression $X'$ is thus transformed to $X' \sim N(0,1)$. 
But now lets say that we are using Poisson regression. I found out through experimentation that subtracting the mean is not useful. If we assume the residuals have a Poisson distribution, then the true mean and true variance of a feature column should be equal. Then we want $X' \sim Poiss(1)$, which is achieved by:
$$X' = \frac{X}{\bar{X}}.$$
From my perspective, this is an effective 'poissonization' for the regression analysis that I am performing; in that Poisson LASSO when features are treated with this transformation gives better results than both untransformed and normalized features.
But I'm trying to find some use of this transformation, and I don't really see it anywhere. I'd like a citation of someone using a transformation like this in literature; or else an explanation of why this is a bad idea.
 A: Technically, it sounds like your method is trying to "unpoisson" the data.
You don't typically see changes-of-variable for a regressor along the lines of what you propose. That's for a couple reasons: 1) You don't need normally distributed regressors, you just want centered/scaled regressors so the L1-penalty is comparing apples-to-apples in terms of effect-size 2) $X/\bar{X}$ has no out of sample validity since the value of $\bar{X}$ is subjective. 3) The variance stabilizing transform $\sqrt{x}$ is known to make more "normally distributed" data out of Poisson values. 4) A log transform is like a square root but has more readily available interpretation.
That said, there's no reason you can't use $\bar{X}$ as a plug-in estimate of the variance. However, to "center-scale" a variable means dividing by the standard error, not the variance, so I would propose the following transformation instead:
$$ X^* = \frac{X-\bar{X}}{\sqrt{\bar{X}}} $$
Alternately, you can just map the quantiles of your Poisson sample onto a standard normal quantile.
As far as evaluating your ideas, it's always good to have an approach in mind. Here's a simulation showing the rate of rejection for Shapiro-Wilk tests of $X/ \bar{X}$ versus $\sqrt{X}$ in samples of Poisson values.
set.seed(123)
p <- replicate(1e5, {
  x <- rpois(100, 10)
  c(
    'xbarx' = shapiro.test(x/mean(x))$p.value,
    'sqrtx'= shapiro.test(sqrt(x))$p.value
  )
})
rowMeans(p < 0.05)

> rowMeans(p < 0.05)
  xbarx   sqrtx 
0.45866 0.34166 

You can see $\sqrt{x}$ rejects the null 34% of the time whereas $x/\bar{x}$ rejects the null 46% of the time: i.e. in a sample of 100 there is more statistical evidence to say the $X/\bar{X}$ is non-normal than the $\sqrt{X}$, putting aside some known issues with the test. 
In summary $X/\bar{X}$ doesn't make the regressor more normal as you say, and normality isn't necessary to begin with.
