Suppose we have a Normal random variable $X$ with the cdf
$$\displaystyle P( X \le x) = \int_{a}^x \frac{\exp \left( -\frac{1}{2} \left(\frac{ t-\mu_1 }{\sigma_1 }\right)^2 \right)}{\sigma \sqrt{2\pi}}dt$$
If we plug this into its own pdf we get a uniformally distributed random variable (Edit: Actually I should write cdf and not pdf)
$$\displaystyle \int_{-\infty}^X \frac{\exp \left( -\frac{1}{2} \left(\frac{ t-\mu_1 }{\sigma_1 }\right)^2 \right)}{\sigma \sqrt{2\pi}}dt \qquad \text{ is uniform on }[0,1]. $$
In fact this holds for any random variable $X$ with sufficiently nice cdf $F$. Plug it into its own cdf and you get a uniformly distributed variable on the range of the cdf.
Added Later:
Since $F(X)$ takes values in $[0,1]$ we only need to consider $x \in [0,1]$ when deriving the cdf of $F(X)$:
$P(F(X) \le x) = P(X \le F^{-1}(X)) = P(X \le y) = F(y) = F (F^{-1} (x)) = x$.
Hence the pdf of $F(X)$ is zero before $x=0$, equal to $1$ after increases linearly to $1$ at $x=1$, and $F(x) =x$ on $[0,1]$. That's a uniform random variable.
But what if we plug $X$ into the pdf of a different normal with different mean and variance?
$$\text{how is } \int_{-\infty}^X \frac{\exp \left( -\frac{1}{2} \left(\frac{ t-\mu_2 }{\sigma_2 }\right)^2 \right)}{\sigma \sqrt{2\pi}} dt \text{ distributed }? $$
I'd expect in general thee things cannot be written out explicitly. But since the normal curve is so well-studied I would not be surprised if the above at least has a name. Does anyone know?
Let $X_1\sim N(\mu_1,\sigma^2_1)$ and $X_2\sim N(\mu_2,\sigma_2^2)\,.$ We know that the CDFs are \begin{align} \mathbb P\{X_i\le x\}=\mathbb P\Big\{\underbrace{\frac{X_i-\mu_i}{\sigma_i}}_{\sim N(0,1)}\le \frac{x-\mu_i}{\sigma_i}\Big\}=\Phi\Big(\frac{x-\mu_i}{\sigma_i}\Big)\tag{1} \end{align} where $\Phi$ is the standard normal CDF: $$ \Phi(z)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^ze^{-y^2/2}\,dy\,.\tag{2} $$ Q. What is the distribution of $X_2$ plugged into the CDF of $X_1\,?$
A. It has the CDF \begin{align} \mathbb P\Bigg\{\Phi\Big(\frac{X_2-\mu_1}{\sigma_1}\Big)\le x\Bigg\}&=\mathbb P\Big\{X_2\le \sigma_1\Phi^{-1}(x)+\sigma_1\mu_1\Big\}\\ &=\Phi\Bigg(\frac{\sigma_1\Phi^{-1}(x)+\sigma_1\mu_1-\mu_2}{\sigma_2} \Bigg)\,.\tag{3} \end{align} Sanity check: when $\sigma_1=\sigma_2$ and $\mu_1=\mu_2$ the RHS collapses to $x$ as it should be the case for a uniform distribution.
In the general case you can now obtain the PDF by taking the derivative of (3) w.r.t. $x\,.$ It does not matter how nasty that expression will be or what name it has but it is doable.