I'm reading about RSA and I have doubts:
1) To choose $e$, this value have to be between $1$ and $\phi=(p-1)(q-1)$, with $\gcd(\phi, e) = 1$, right?
2) In my example, $p = 139$ and $q = 491$. So $n = 68249$ and $\phi = 67620$. Assuming my point 1 is correct, $e$ can be $67619$. But $d$ can be $67619$ too, because $67619^2 \equiv 1 \mod 67620$.
Is this reasoning incorrect?
1) To choose $e$, this value have to be between $1$ and $\phi=(p-1)(q-1)$, with $\gcd(\phi, e) = 1$, right?
No; $e$ can be any value that's relatively prime to both $p-1$ and $q-1$ (or equivalently, relatively prime to $\operatorname{lcm}(p-1, q-1)$). There may be little point in choosing an $e$ larger than $n$; however there's no specific reason it wouldn't work.
2) [Paraphrased] What if I pick an $e$ with $e^2 = 1 \bmod (p-1)(q-1)$; and set $d = e$. Would it work?
Well, yes, it would work, in the sense that the protocol will encrypt and decrypt messages just fine. It wouldn't work in the sense that it wouldn't be secure (even if you used more realistic sizes of $p$ and $q$).
Yes, math behind this work. But it seems that you are asking, can both public and private exponent be the same value? Answer is no, because you jeopardize the whole system. By choosing equal exponents, you create two identical keys. And if some eavesdropper steals public key, he can decrypt message.
Let $e\ne 1,-1$ be the public exponent such that $e^2=1 \pmod {\phi(N)}$, so we have:
$$(e-1)\cdot (e+1)=0 \pmod {\phi(N)}$$
then we can compute $\phi(N)$ and factor $N$.
