Number field determined by its prime factorisations Is a number field uniquely determined by the factorisations of the rational primes? In other words, are there two number fields $E, E'$ such that the there is a bijection between the primes above $p$ with matching inertia degree and ramification degree?
This is the case if $E/\mathbb{Q}$ is Galois, as was discussed here. https://mathoverflow.net/questions/186255/is-a-number-field-uniquely-determined-by-the-primes-which-split-in-it
This answer also mentions that there are non-isomorphic number fields such that the primes not dividing the discriminant have the same factorisation, an example can be found in Cassels-Fröhlich Algebraic Number Theory Exercises 6.3 and 6.4, where they state that

The same primes $p$ are ramified in $E$ as in $E'$, and for non-ramified $p$ the decomposition of $p$ in $E$ and $E'$ is the same, in the sense that the collection of degrees of the factors of $p$ in $E$ is identical with the collection of degrees of the factors of $p$ in $E'$, or equivalently, in the sense that $A/pA \approx A'/pA'$, where $A$ and $A'$ denote the rings of integers in $E$ and $E'$ respectively.

The wording of this statement suggests that the factorisation of ramified primes in $E$ and $E'$ does not match. To me this suggests that such an example may not exist.
Consider that $\mathrm{Spec}(A) \to \mathrm{Spec}(\mathbb Z)$ is a branched covering, the result would follow from some kind of algebro-geometric statement about the uniquess of a branched covering of schemes under the assumption that the fibres are isomorphic. My intuition is that this is a plausible statement, but it seems non-trivial enough that a counterexample could reasonably exist.
 A: Let $K$ be a number field, for each $p$, we can associate $$A_{p,K} = \{(e_1,f_1),\cdots,(e_g,f_g)\}\qquad $$ which denotes ramification indices, inertial degrees of each prime lying above $p$. I believe OP is asking if for two $K,K'$, we have $A_{p,K} = A_{p,K'}$, does it imply that $K, K'$ are isomorphic?
The answer is negative, one concrete example is given below.

Set $B_{p,K} = \{f_1,\cdots,f_g\}$ for list of inertial degrees. $B_{p,K} = B_{p,K'}$ for all $p$ is equivalent to $K,K'$ being arithmetically equivalent (AE) (i.e. having the same zeta function). AE fields have same Galois closure. Given $K$, existence of other AE field(s) depends entirely on the Galois group of $K$.
The smallest $[K:\mathbb{Q}]$ for which $K$ can have AE sibling is $[K:\mathbb{Q}]=7$, with Galois group $\text{GL}_3(\mathbb{F}_2)$. They satisfy $B_{p,K'} = B_{p,K}$ for all $p$, so $A_{p,K}=A_{p,K'}$ for unramified $p$. For ramified $p$, $A_{p,K}, A_{p,K'}$ may or may not be the same, it now depends on $K,K'$.
Not all degree $7$ number fields with this Galois group will satisfy $A_{K,p} = A_{K,p}$ for ramified $p$, but one pair do is defined by
$$f(x) = x^7 - 3x^6 + 3x^5 + 3x^4 - 13x^3 + 17x^2 - 10x + 1$$
and $$g(x) = x^7 - 2x^5 - 4x^4 - x^3 + 5x^2 + 5x + 1$$
only ramified at $p=3803$, and $A_{p,K} = A_{p,K'} = \{(2,2),(1,2),(1,1)\}$.

Since many invariants of number fields defined by $f,g$ are same, it would be interesting to show they are indeed not isomorphic.
One way is just to look at the defining polynomial on LMFDB, which is output of PARI's polredabs, two isomorphic fields always have same defining polynomial on LMFDB.
Another more self-contained approach is to compute the resultant $$h(X) = \text{res}_x(f(x), g(X-x)) = \prod_{1\leq i,j\leq 7}(X-\alpha_i - \beta_j) \in \mathbb{Z}[X]$$ where $\alpha_i, \beta_j$ are roots of $f,g$ respectively. Note that $h$ has degree $7^2$. If $K,K'$ were isomorphic, then $h$ would have a factor of degree $7$ over $\mathbb{Q}$ (because some $\alpha_i$ would be in $\mathbb{Q}(\beta_j)$). Some computations show degrees of irreducible factors of $h$ are $21,28$, so $K, K'$ are not isomorphic.
