Not sure if "linear transformation" is the correct terminology, but...
Let $X$ be a random variable with a normal distribution $f(x)$ with mean $\mu_{X}$ and standard deviation $\sigma_{X}$: $f(x) = \frac{1}{\sigma_{X}\sqrt{\tau}}\exp{\left[\frac{-1}{2}\left(\frac{x-\mu_{X}}{\sigma_{X}}\right)^2\right]}$
(Here, $\tau=2\pi$)
Let $Y$ be a random variable defined by the linear transformation $Y = u(X) = aX+b$
Let $v(y) = u^{-1}(y) = \frac{y-b}{a}$. Then $v^\prime(y) = \frac{1}{a}$.
Prove: $Y$ is normally distributed, with density function $g(y)=f\bigl(v(y)\bigr)\,\bigl|v^\prime(y)\bigr|$ $= \frac{1}{\sigma_{X}\sqrt{\tau}}\exp{\left[\frac{-1}{2}\left(\frac{\frac{y-b}{a}-\mu_{X}}{\sigma_{X}}\right)^2\right]}\,\left|\frac{1}{a}\right|$ $= \frac{1}{\left|a\right|\sigma_{X}\sqrt{\tau}}\exp{\left[\frac{-1}{2}\left(\frac{y-(a\mu_{X}+b)}{a\sigma_{X}}\right)^2\right]}$ with mean $\mu_{Y} = a\mu_{X}+b$ and standard deviation $\sigma_{Y} = \left|a\right|\sigma_{X}$.