The relationship between the pressure at the base of a container and the normal reaction on the container This is NOT a homework question. The question here is just an example.

In this question, after both the liquids are mixed and a homogenous mixture is created, I found the effective density of the mixture and found that the pressure at the base increases by 1/2(rho)gh.
This was calculated by taking a smaller cylinder with area a inside the system and as the fluid inside this imaginary cylinder is in equilibrium, I took the pressure at the bottom into the area of that smaller cylinder as the weight of the fluid.

The normal reaction on the container by the ground remains the same obviously as there is no change in the mass of the system.
Now what I think is that the pressure on the base × base area = normal force
As the base area hasn't changed but the pressure has, the normal force also changes? Which is clearly wrong.
I have clearly misunderstood something here and would like to know if the pressure at the bottom of a container and normal force are related and how.
 A: To get

the pressure at the base increases by 1/2(rho)gh

presumably you've found that the new density is $\frac{3\rho}{2}$ and done this:
Original pressure is $$2H\rho g + H (2\rho) g = 4H\rho g\tag 1$$
and the new pressure is
$$2H\frac{3\rho}{2} g + H\frac{3\rho}{2}  g = \frac{9}{2}H\rho g\tag 2$$
But you should divide the contribution of the top cylinder, to the pressure at the base, by 2 due to the increase of area, as the force due to the weight of the top cylinder spreads evenly throughout, (first term) and change equations 1) and 2) to 3) and 4)
$$H\rho g + H (2\rho) g = 3H\rho g\tag 3$$
$$H\frac{3\rho}{2} g + H\frac{3\rho}{2}  g = 3H\rho g\tag 4$$
so A) and B) seem ok.
A: 
Now what I think is that the pressure on the base × base area = normal force

If you mean the internal pressure in the vessel, then no.  The vessel has rigid walls and the pressure inside is not necessarily related to the pressure outside.
Imagine a gas cylinder with a movable piston.  We can depress the piston to increase the pressure inside without adding mass.  But the normal force from the ground is unaltered.
If the fluid pressure on the bottom of the vessel increases, but the normal force does not, then some other force must be changing so that the bottom does not accelerate.  This force comes from the walls of the cylinder.
The extra area of the bottom cylinder that is not under the upper cylinder is under internal pressure.  Before the mixing the fluid is pushing up on this area with pressure $2\rho g H$, but after the mixing the pressure is $3 \rho g H$.  This extra pressure creates an upward force that counteracts the force from the increased downward pressure on the base.
