Yes, it's certainly true. We say that $f(n)\in O(g(n))$ if there exists a constant $C_1$ such that $\lvert{f(n)}\rvert < C_1\lvert g(n) \rvert$ for sufficiently large $n$, and similarly $g(n)\in O(f(n))$ if there exists $C_2$ such that $\lvert{g(n)}\rvert < C_2\lvert f(n)\rvert$ for sufficiently large $n$. These both hold if there are constants $A$ and $B$ such that $A\lvert g(n) \rvert < \lvert{f(n)}\rvert < B\lvert g(n) \rvert$ for sufficiently large $n$, or equivalently such that $ A < \left\lvert{\frac{f(n)}{g(n)}}\right\rvert < B. $ In this case we say that $f(n)\in\Theta(g(n))$. The relation is symmetric and transitive, and obviously reflexive, so it defines an equivalence relation on the set of functions.