@srestha do you know the answer ? I think both should be true, cause all upper bounds to $O(n)$ can be upper bounded by $O(n^2)$.
I'm not sure if it makes sense but for all the functions that we pick for left hand side, we can always pick another asymptotically greater function on the right hand side, hence making the Statement valid. Basically what I'm saying is, $O(n) = O(n^2)$ means set of all functions that are upper bounds to $n$ can be upper bounded by set of functions that are upper bound to $O(n^2)$
$\Theta(n)$ won't work here, and if we flip the functions on left hand side and right hand side, I think $\Omega(n)$ should work too.