Are there two non-isomorphic number fields with the same degree, class number and discriminant? If so, do people expect certain invariants (regulator, # of complex embeddings, etc) to fully 'discriminate' between number fields?
 A: Here's a degree three example with minimal discriminant: the two cubic fields given by $x^3-21x-35$ and $x^3-21x+28$ 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 non-isomorphic fields of that discriminant.
A: 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. 
A: 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 non-isomorphic (e.g. the regulators distinguish them, the splitting fields distinguish them...).
A: 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 $x^3-18x-6$, $x^3-36x-78$ and $x^3-54x-150$. The resulting fields are non-isomorphic cubics with class number one and discriminant $22356$.
A: 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.
