In a column of fluid does density vary? In a column of fluid does density vary?
 A: It depends on the fluid.  
Consider, for example, an ideal gas at fixed temperature near the surface of the Earth.  Does the density vary in such a column?  Yes.
Let's investigate as follows.  Imagine that the column is in the $z$-direction and has cross-sectional area $A$.  Let $z=0$ at the ground.  Consider a small, vertical "piece" of the column between height $z$ and $z+\Delta z$.  The force pushing up on the bottom of this piece due to the air below it is the pressure $P(z)$ at the bottom times the cross sectional area $A$ of the piece.
\begin{align}
  F_\mathrm{top} = P(z)A
\end{align}
Similarly, the force pushing down on top due to the air above is
\begin{align}
  F_\mathrm{bottom} = -P(z+\Delta z)A.
\end{align}
If the column is in equilibrium, these forces along with the weight of the piece will sum to zero -- they will balance.  Since the mass of the piece is it's volume times its density, and since $\Delta z$ is taken small, the density can be taken to be the density $\rho(z)$ at its bottom which gives a mass $A\Delta z \,\rho(z)$.  Therefore, it's weight is
\begin{align}
  W = -A\Delta z\rho(z) g
\end{align}
Summing up all the forces and setting them to zero, namely 
\begin{align}
  F_\mathrm{top} + F_\mathrm{bottom} + W = 0
\end{align}
gives
\begin{align}
  P(z)A-P(z+\Delta z)A-A\Delta z\rho(z) g = 0
\end{align}
which can be rearranged to give
\begin{align}
  \frac{P(z+\Delta z) - P(z)}{\Delta z} = -\rho(z) g
\end{align}
If one takes the limit $\Delta z \to 0$, then the left hand side becomes $P'(z)$, namely the derivative of the pressure with respect to $z$, so we obtain a relationship between the derivative of the pressure and the density
\begin{align}
  P'(z) = -\rho(z) g \tag{$\star$}
\end{align}
On the other hand, the ideal gas Law tell us that
\begin{align}
  P(z) = \frac{\rho(z)}{m}kT
\end{align}
where $m$ is the mass per molecule, $k$ is Boltzmann's constant, and $T$ is the temperature of the gas.  Combining this with $(\star)$ gives the following differential equation for the density:
\begin{align}
  \rho'(z) = -\frac{mg}{kT} \rho(z)
\end{align}
and therefore, the density as a function of height $z$ above the ground is
\begin{align}
  \boxed{\rho(z) = \rho(0) e^{-\frac{mg}{kT} z}}
\end{align}
In other words, in this simplified model, the density of an ideal gas decreases exponentially with height.
On the other hand, a fluid like water does not obey the ideal gas law, and in fact, water can be well-approximated as an incompressible fluid, so the density of a column of water would not vary significantly with height.
