I'm talking about the set theory definition of $O(f(n))$, which is the set of all functions $g$ such that $O(f(n))=O(g(n))$.
What is the cardinality of $\{O(f(n)):f\in\mathbb{R}^\mathbb{R}\}$?
We can denote this cardinality $\kappa$. We know that $\kappa\geq\mathfrak{c}$ (because for every real number $x$, $O(x^n)$ is distinct).
We also know that $\kappa\leq\mathfrak{c}^\mathfrak{c}=2^\mathfrak{c}$, because the set it is the cardinality of has an injection onto $\{f:f\in\mathbb{R}^\mathbb{R}\}=\mathbb{R}^\mathbb{R}$.
So, we have upper and lower bounds. Assuming GCH holds for $\aleph_0$ and $\aleph_1$ (i.e. $\mathfrak{c}=\aleph_1$ and $2^\mathfrak{c}=\aleph_2$), then it is shown that $\aleph_1\leq\kappa\leq\aleph_2$, meaning it is either $\mathfrak{c}$ or $2^\mathfrak{c}$.
However, if ZFC can prove that $\mathfrak{c}<\kappa<2^\mathfrak{c}$, it would immediately disprove GCH. Assuming ZFC is consistent and $\mathfrak{c}<\kappa<2^\mathfrak{c}$, then ZFC can't prove this is true, and clearly can't give an exact cardinality for $\kappa$.
So, I would encourage those solving this problem to try and either prove $\kappa=\mathfrak{c}$ or $\kappa=2^\mathfrak{c}$, or your won't get very far.