It is possible for nonisomorphic fields to have the same discriminant, and even to be arithmetically equivalent, meaning with the same Dedekind zeta function. Previous questions which discussed this are "Number fields with same discriminant and regulator?" and "Are there two nonisomorphic number fields with the same degree, class number and discriminant?". The usual way this comes about involves fields with isomorphic subfields. I would find it useful if the discriminant sufficed to characterize number fields whose splitting field had a simple Galois group, so that there were no subfields. So the question is, can there be two nonisomorphic fields with the same degree, the same discriminant and the same simple Galois group for the splitting field?
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 $t1$ distinct primes that are congruent to 1 mod 3, then there are exactly $2^{t1}$ 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 nonisomorphic 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 3cycle 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.

$\begingroup$ I'm really just interested in the nonabelian case, if someone has an answer for that, but I didn't see how to use the conductor discriminant formula for the abelian case. That it doesn't work then is not encouraging for nonabelian groups. $\endgroup$ – Gene Ward Smith Sep 17 '12 at 23:20
I just came upon this (old?) thread. Here is a handy way to produce lots of nonAbelian, nonisomorphic extensions of Q with the same field discriminant:
Let K3/Q be a nonAbelian (resp Abelian) cubic field. Let h3 be the size of the elementary 2class group of K3. Then Heilbronn (On the 2classgroup of cubic fields, in: Studies in Pure Math}., 1971) showed that there is a (h31) (resp (h31)/3) quadruplets of conjugate quartic fields K4 such that disc(K4)=disc(K3) and that the Galois closure of K4 contains K3.
So if you pick a K3 with large elementary 2rank, then you will get lots of nonisomorphic K4 with Galois group S4 or A4, depending on whether K3 is nonAbelian or otherwise.
There is also a similar (and old) theorem of Hasse for 3rank of quadratic fields.