I came across this question recently and it has me scratching my head. For a standard normal random variable $X$, with density $\Phi(X)$, let $Y = \Phi(X)$. What is the distribution of Y?
I had two thoughts, the first is to use just a standard transformation technique and brute force it, the second is to utilize the delta-method, but since the delta-method is really for asymptotic relationships I decided against it.
So brute force method it is!
The transformation function is
$$Y \overset{set}{=} \frac{1}{\sqrt{2\pi}} \exp(-\frac{X^2}{2})$$
Get $X$ in terms of $Y$,
$$\Rightarrow X = \sqrt{\log \left ( \frac{1}{2\pi Y^2}\right)}$$
Calculate Jacobian
$$\frac{\partial X}{\partial Y} = -\left ( \log \left ( \frac{1}{2\pi Y^2}\right )\right )^{-\frac{1}{2}} \frac{1}{Y}$$
$$ = \left ( \log \left ( 2\pi Y^2\right )\right )^{-\frac{1}{2}} \frac{1}{Y}$$
Find density of $Y$
$$\Rightarrow \frac{1}{\sqrt{2\pi}} \exp \left (-\frac{\sqrt{\log \left ( \frac{1}{2\pi Y^2}\right)}^2}{2} \right ) \left ( \log \left ( 2\pi Y^2\right )\right )^{-\frac{1}{2}} \frac{1}{Y}$$
$$ =\frac{1}{\sqrt{2\pi}} \exp \left (\frac{\log \left (2\pi Y^2\right)}{2} \right ) \left ( \log \left ( 2\pi Y^2\right )\right )^{-\frac{1}{2}} \frac{1}{Y}$$
$$ =\frac{1}{\sqrt{2\pi}} \exp \left ( \log \left ( \sqrt{2\pi Y^2}\right)\right ) \left ( \log \left ( 2\pi Y^2\right )\right )^{-\frac{1}{2}} \frac{1}{Y}$$
$$ =\frac{1}{\sqrt{2\pi}} \left ( \sqrt{2\pi} Y\right)\left ( \log \left ( 2\pi Y^2\right )\right )^{-\frac{1}{2}} \frac{1}{Y}$$
$$ \Rightarrow f_Y(y) =\log \left ( 2\pi Y^2\right )^{-\frac{1}{2}}, \text{for } 0<Y<1$$
Now, I am doubtful this is the right approach, especially because the density of $Y$ does not map on to the real numbers. My guess is that there is a trick I do not know about, and/or I missed something pretty big I didn't consider. Thoughts and suggestions appreciated :)
The correct density for $Y$ is $$f_Y(y) = 2 \left(-\log (2 \pi y^2)\right)^{-1/2}.$$ The first problem with your result is that when $2\pi y^2 < 1$, its logarithm is negative and the square root is not real-valued. The second problem is that you are missing a factor of $2$ because the transformation is not one-to-one. That said, your basic approach is correct.
Under the transformation $$Y = \frac{1}{\sqrt{2\pi}} e^{-X^2/2},$$ for $X \in \mathbb R$, we have the support $Y \in (0, (2\pi)^{-1/2}]$. Therefore, we have $$F_Y(y) = \Pr[Y \le y] = \Pr\left[X^2 \ge -2 \log (y\sqrt{2\pi})\right] = \Pr\left[X^2 \ge -\log (2\pi y^2)\right].$$ Since $X$ is standard normal, this is simply $$F_Y(y) = 2 F_X\left(-\left(-\log (2\pi y^2)\right)^{1/2}\right).$$ It follows that $$f_Y(y) = 2 f_X\left(-\left(-\log (2\pi y^2)\right)^{1/2}\right) \cdot \frac{1}{y (- \log (2\pi y^2))^{1/2}}= 2\left(-\log (2\pi y^2)\right)^{-1/2}$$ as claimed.