If so, do people expect certain invariants (regulator, # of complex embeddings, etc) to fully 'discriminate' between number fields?
John Jones has computed tables of number fields of low degree with prescribed ramification. Though the tables just list the defining polynomials and the set of ramified primes, and not any other invariants, it's not hard to search them to find, e.g., that the three quartic fields obtained by adjoining a root of $x^4  6$, $x^4  24$, and $x^4  12x^2  16x + 12$ respectively all have degree $4$, class number $1$, and discriminant $2^{11} \cdot 3^3$. On the other hand these three fields are nonisomorphic (e.g. the regulators distinguish them, the splitting fields distinguish them...).
An example (this is exercise 21 in Chapter 2.2 of Borevich and Shafarevich) is given by the three cubic fields obtained by adjoining a root of $x^318x6$, $x^336x78$ and $x^354x150$. The resulting fields are nonisomorphic cubics with class number one and discriminant $22356$.

8$\begingroup$ This is the smallest discriminant shared by three nonisomorphic totally real cubic fields. $\endgroup$ – KConrad Jun 25 '10 at 5:29
This isn't quite your question, but there are nonisomorphic number fields with the same zeta function. They are built in this way: Take a Galois extension $K/Q$, with Galois group $G$. Find subgroups $H_1$ and $H_2$ so that $G/H_1$ and $G/H_2$ are nonisomorphic as sets with $G$action but $C^{G/H_1}$ and $C^{G/H_2}$ are isomorphic as vector spaces with a $G$action. Then look at the fixed fields of $H_1$ and $H_2$.
Because there is a lot of freedom in this construction, I would guess that you can manipulate it to avoid detection by any of the easy invariants you mention.

4$\begingroup$ If you choose them so that the isomorphism k^{G/H_1} = k^{G/H_2} works over any field (there are examples in my undergrad thesis), the class numbers have to be the same. $\endgroup$ – Ben Webster♦ Oct 30 '09 at 21:27
Here's a degree three example with minimal discriminant: the two cubic fields given by $x^321x35$ and $x^321x+28$ both have discriminant $3969=63^2$, both are Galois, both have class number 3. The regulator of the first is about 4.20169 and that of the second is about 12.59419. This is the smallest (positive) discriminant of a degree 3 field to have two nonisomorphic fields of that discriminant.

$\begingroup$ It's easier (well, if you were on the proverbial desert island without a computer) to check they are not isomorphic by looking at prime splitting: 2 is inert in the first and splits in the second. $\endgroup$ – KConrad Jun 25 '10 at 5:31

1$\begingroup$ Indeed. Hopefully desert islands will have free wifi some time soon. $\endgroup$ – Rob Harron Jun 25 '10 at 7:21

1$\begingroup$ There are examples with smaller absolute discriminant: the "pure" cubic fields obtained by adjoining cube roots of 6 and 12. Both have discriminant 972 and class number 1, but they're not isomorphic. Pure cubics give lots of examples like this  I was able to find 5 fields with common discriminant and class number. $\endgroup$ – Ari Shnidman Mar 16 '12 at 6:10

$\begingroup$ @Ari: yeah, sorry, I meant of positive discriminant. $\endgroup$ – Rob Harron Sep 17 '12 at 19:31
I recently heard a fascinating talk by Drew Sutherland on work he's done on this sort of question. Rather than trying to summarize his talk, here is a link to the slides for his talk. He also indicated that he'll be posting an article soon on the ArXiv, but it's not quite ready yet.