Lognormal standardisation Suppose we have a standardized normal $Z_1 \sim \mathcal{N}(0,1)$, and take the associated lognormal $X = e^{Z_1}$.
Suppose that with have a non-standardized normal $Z_2 \sim \mathcal{N}(\mu,\sigma^2)$ and the associated lognormal $Y = e^{Z_2}$
Is there somewhere an expression in terms of $\mu,\sigma$ for standardisation parameters $a,b \in \mathbb{R}\times\mathbb{R}_{+}$ such that, in distribution, 
$$Y = aX+b \text{ ?}$$
 A: Unless $\sigma=1,$ no such linear transformation exists.  This is an immediate consequence of the one-to-one correspondence between $\sigma$ and the skewness of a lognormal variable, because linear transformations do not change the magnitude of the skewness and lognormal variables always have positive skewness.

Consider the skewness of $Y=\exp(Z_2),$ given by
$$\beta_3(Y)= \left(\exp(\sigma^2) + 2\right)\sqrt{\exp(\sigma^2)-1}.$$
The skewness of any variable is the third moment of its standardized version of the variable.  Because linear transformations with positive slopes do not change the standardization, they leave the skewness unchanged.  Thus, from the two expressions $aX+b=Y=\exp(Z_2)$ and noting $a\gt 0$ we obtain
$$\left(e^1+2\right)\sqrt{e^1-1} = \beta_3(aX+b) = \beta_3(Y) = \left(e^{\sigma^2/2} + 2\right)\sqrt{e^{\sigma^2/2} - 1}.\tag{1}$$
The claim is that $\sigma=1$ is the only (positive) solution to this equation.  To see why that is so, for $t\gt 0$ define
$$f(t) = \sqrt{t(t+3)^2}.$$
Writing $\lambda = \exp(\sigma^2)-1$ and $\rho = \exp(1)-1$ (both of which are positive), equation $(1)$ asserts
$$f(\lambda) = f(\rho).$$
However, as a function defined on the positive numbers, $f$ is one-to-one.  This is readily seen by noting that the derivative
$$\frac{d}{dt} t(t+3)^2 = 3(t^2 + 4t + 3)$$
can never be zero for $t\gt 0,$ implying $f^2$ is a positive one-to-one function, whence so is $f.$  Consequently $\lambda=\rho,$ showing that $\sigma=1$ is the only solution.
(On the other hand, when $\sigma=1,$ $Z_2 = Z_1 + \mu$ and therefore 
$$Y = \exp(Z_2) = \exp(Z_1+\mu) = e^\mu \exp(Z_1) = e^\mu X + 0$$
indeed is a linear transformation of $X$ with $a=e^\mu,b=0.$)
