My answer to one of those previous questions provides an example of two Galois cubic fields of the same discriminant. The cyclic group of order three is simple.
Added: Just to give you an idea, if you look in Cohen's book A course in computational algebraic number theory, Theorem 6.4.6 says that if $e$ is the product of 9 times $t-1$ distinct primes that are congruent to 1 mod 3, then there are exactly $2^{t-1}$ cyclic cubic fields of discriminant $e^2$. There's a similar statement for $9\nmid e$.
And more: So using sage/pari, I just found the following example of two non-isomorphic quintic $A_5$-extensions with the same discriminant. Take $f=x^5 - x^2 - 2x - 3$ and $g=x^5 - x^4 + 5x^3 - 3x^2 + 4x - 3$. The associated fields (and the polynomials themselves) have discriminant 243049 (and 1 real place), which is a square. Hence, the Galois group is $A_5, D_5$, or $C_5$. Their factorization mod 3 are $x(x + 1)(x^3 + 2x^2 + x + 1)$ and $x (x + 2) (x^3 + 2x + 2)$, respectively. Hence, both of their Galois groups contain a 3-cycle and hence must be $A_5$.
In fact, these both have class number 1. However, 5 is inert in the second field, but splits into primes of degree 1,1, and 3 in the first field. Their regulators are 10.7855048337065 and 7.68746944439494, respectively.
