I am looking for a closed form relation between $x_1$ and $x_2$ that equates two normal CDFs of the same mean but different standard deviation:
$F(x_1;\mu,\sigma_1)=F(x_2;\mu,\sigma_2)$
Also, as a follow up I am looking for a similar relation between $x_1$ and $x_2$ for the following:
$(1-F(x_1;\mu,\sigma))^N=1-F(x_2;\mu,\sigma/\sqrt{N})$
Basically, I want a relation that for some value of $x_1$ it gives me $x_2$: $x_2=f(x_1,\mu,\sigma,N)$
For your first question, we have that $F(x_1;\mu, \sigma_1) = F(\frac{x_1-\mu}{\sigma_1};0,1)$, so $(F(\frac{x_1-\mu}{\sigma_1};0,1) = F(\frac{x_2-\mu}{\sigma_2};0,1)$. Since $F(x,0,1)$ is injective, this means that $\frac{x_1-\mu}{\sigma_1} = \frac{x_2-\mu}{\sigma_2}$, or that $\frac{x_1}{\sigma_1} = \frac{x_2}{\sigma_2}$.
The second question is considerably more difficult, because the $N$th power on the left side makes it more impossible to reduce to a simple equality of normal CDFs.
