Scaling of random variables Note: I will use $\mathbb{P}\{X \in dx\}$ to denote $f(x)dx$ where $f(x)$ is the pdf of $X$.
While doing some homework, I came across a fault in my intuition. I was scaling a standard normally distributed random variable $Z$. Edit: I was missing the infinitesimals $dx$ and $dx/c$, so everything works out in the end. Thank you Jokiri!
$$\mathbb{P}\{cX \in dx\} = \frac{e^{-x^2 / 2c^2}}{\sqrt{2\pi c^2}}$$
while
$$\mathbb{P} \left\{X \in \frac{dx}{c}\right\} = \frac{e^{-(x/c)^2/2}}{\sqrt{2\pi}}$$

Could anyone help me understand why the following equality doesn't hold? Edit: it does, see edit below
  $$\mathbb{P}\{cX \in dx\} \ne  \mathbb{P} \left\{X \in \frac{dx}{c}\right\}$$

I have been looking around, and it seems that equality of the cdf holds, though:
$$\mathbb{P}\{cX < x \} = \mathbb{P}\left\{X < \frac xc\right\}.$$
Thank you in advance!

This question came out of a silly mistake on my part. Let me attempt to try to set things straight.
Let $Y = cX$. Let $X$ have pdf $f_X$ and $Y$ have pdf $f_Y$.
$$\mathbb{E}[Y] = \mathbb{E}[cX] = \int_{-\infty}^\infty cx\,f_X(x)\mathop{dx}
=\int_{-\infty}^\infty y\, f_X(y/c) \frac{dy}{c}$$
So, $f_Y(y) = \frac 1c f_X(y/c)$.
Thank you for the help, and sorry for my mistake.

Setting aside rigour and following your intuition about infinitesimal probabilities of finding a random variable in an infinitesimal interval, I note that the left-hand sides of your first two equations are infinitesimal whereas the right-hand sides are finite. So these are clearly wrong, even loosely interpreted. They make sense if you multiply the right-hand sides by the infinitesimal interval lengths, $\mathrm dx$ and $\mathrm dx/c$, respectively, and then everything comes out right.
