In NSA (ZFC+IST), what can we say about generators for $\mathbb{Z}/\nu\mathbb{Z}$ for unlimited $\nu$?

You are asking if the multiplicative group of $\mathbb{Z}/\nu\mathbb{Z}$ is cyclic, and this is true: $\mathbb{Z}/\nu\mathbb{Z}$ is a finite field, and multiplicative groups of finite fields are cyclic, nothing stops you carrying out the usual proof in IST.

Consider the set of primes $p$ for which 2 is a generator of the multiplicative group mod $p$. This is conjectured to be infinite, in which case it contains a nonstandard element (that infinite sets contain nonstandard elements is proved in Robert's book, indeed it's Basic Principle number 1 on the inside cover of my copy). Better, according to an answer to this MO question the set of primes for which $2,3$ or $5$ is a primitive root is known to be infinite, so there are illimited primes whose multiplicative groups are generated by $2,3$ or $5$.

The number of generators of the multiplicative group mod $\nu$ is $\phi(\nu-1)$. Since $\nu-1$ is illimited, so is $\phi(\nu-1)$ (the prime factorization of $\nu-1$ must have either an illimited prime or a limited prime to an illimited power or contain an illimited number of limited distinct primes, so use multiplicitivity of $\phi$).

Lemma: a set of natural numbers whose size is illimited must contain an illimited element. Proof: the set is finite, so has a maximum. If the maximum was standard then the set would be contained in a standard interval, so have standard size.

We can conclude there is always an illimited primitive root.

It is a theorem that there exists a (standard!) constant $C$ such that there are infinitely many primes $p$ for which the least positive primitive root has size $C \log p$. Thus there is an illimited prime $\nu$ satisfying this, so all the elements of $[0,\nu-1]$ which generate the multiplicative group are illimited. (This is exactly André Nicolas' comment above).