I may be wrong, but it seems to me that the function defined by:

$f(n) = left{begin{array}{rl}1 & text{if }n=1\2^{2^{leftlfloor log_2log_2 n rightrfloor+1}}&text{otherwise}end{array}right.$

satisfy the conditions.

To explain why, the values taken by $f(n)$ will be $1, 4, 4, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 256, 256, …$

The idea is that when $n=2^{2^k}$, then $f(n) = 2^{2^{k+1}} = n^2$, and when $n=2^{2^k}-1$, then $f(n) = 2^{2^{k}} = n +1$. That way, you will guarantee the 5 conditions: $(O(n^2) setminus o(n^2)) cap (Omega(n) setminus omega(n))$ and increasing.