Assume we have a somehow unorthodox implementation of RSA whereby:
$p$ and $q$ are chosen primes of length $n/2$ where $n$ is the number of bits desired in $N=p\,q$ and $$\begin{align} \phi &= (p-1)(q-1)\\ d &= p\\ e &= d^{-1}\bmod \phi \end{align}$$
I can see that there is a weakness here but I'm unsure of how it translates. Playing around a bit I see that I end up with:
$$e = \frac{\phi j + 1}p $$
Further assuming that $\frac{1}p$ can be considered infinitesimal: $$e ≈ j\left(q\left(1+\frac{1}p\right) -1\right)$$
So basically: $$e ≈ j(q-1)$$
Is that right? Am I missing something?