Let's take a relatively strong axiomatic system, call it $A$. By Godel's Incompleteness Theorem, there are infinitely many true statements expressible in $A$ that cannot be proven. Let's call the set of all truths in $A$ that are provable, $T$. And then all nonprovable truths $NT$.
If we have another axiom system where all the axioms are the negations of $A$'s axioms, i.e $\neg A$, the set of provable theorems is trivially $\neg T$. My question is if the same applies to $NT$, if the set of all unprovable statements in $\neg A$ is $\neg NT$.
In simpler terms, are all unprovable statements in the negation of an axiomatic system the negation of an unprovable statement in the original system?
Your second paragraph is incorrect. We can have a consistent theory $A$ such that $\neg A$ is inconsistent: this means that $\neg A$ proves everything, not just the negations of the theorems of $A$. Similarly, if $A$ is inconsistent but $\neg A$ is consistent, then everything is a negation of a theorem of $A$, so $\neg A$ does not prove all negations of theorems of $A$. There are plenty of further examples. For instance, take $p\in A$ and $\neg q\in A$. Then the sentence "$p\vee q$" is a theorem of both $A$ and $\neg A$, even if it isn't a tautology.
So in general there is no obvious relationship between the theorems of $A$ and the theorems of $\neg A$. Similarly, the answer to the question you ask is "no."