Show that a Linear transformation does not change shape of a distribution It is easy to show how a linear transformation affects the mean or the variance of a distribution. It is easy to find over the internet that a linear transformation does not change the shape of a distribution.
My question is, why a linear transformation does not change the shape of a distribution? Is it an easy to show proof?

Showing this requires a definition of "shape". All that comes to mind to me is what the distribution looks like if it is centered and scaled. This is straightforward when the mean exists and variance is finite, so I will proceed with such an assumption.
Let $X$ be a random variable with distribution function $F_X(x)$. Define a new random variable, $Z$, as a function of $X$.
$$
Z(X) = \dfrac{X-\mathbb E \big[X\big]}
{\sqrt{var(X)}}
$$
Now let $T(X) = a + bX$ be a linear transformation with $b \ne 0$.
$$
\mathbb E\big[ T(X)\big] = a + b\mathbb E\big[ X \big]\\
var(T(X)) = b^2var(X)
$$
Let's apply that same logic of centering and scaling $T(X)$.
$$
Z(T(X)) = \dfrac{
T(X) - \mathbb E\big[ T(X)\big]
}{
\sqrt{var(T(X))}
} 
= \dfrac{
a + bX -(a + b\mathbb E\big[ X \big])
}{
\sqrt{b^2var(X)}
}
=\dfrac{
bX-b\mathbb E\big[ X \big]
}{
b\sqrt{var(X)}
} \\= \dfrac{X-\mathbb E \big[X\big]}
{\sqrt{var(X)}} = Z(X)
$$
