What is the property of RSA where N=e? In RSA, suppose we know that $e=N$ and we are given the value of $e$. 
($N =  p\cdot q$ for some large primes $p$ and $q$; $\gcd(e, \varphi(N) = 1)$

Can we calculate $d$ ($d = e^{-1} \mod \varphi(N)$) without factoring $N$?
No, $e=N$ is no less secure than any other value of $e&gt;1$.

In fact, we can make a stronger statement; if setting $e=N$ allows us to compute the corresponding $d = N^{-1} \mod \phi(N)$, then we can factor $N$ (and hence we can find the $d$ corresponding to any $e$).  This is due to the fact that knowledge of any nontrivial pair $(d,e)$ with $d \cdot e = 1 \mod \phi(N)$ allows us to efficiently find the factorization of $N$.

We don't use $e=N$, not because of security reasons, but because of efficiency. Setting $e=N$ makes the public key operation (encryption, signature verification) far more expensive than it would be if we set $e$ to a small value; given that a large $e$ typically would have no advantage to make up for that, we generally don't use large $e$.

Historical note: when Clifford Cox initially devised his varient of RSA (which wasn't declassified until well after RSA was published), he actually had $e = N$.