The density of a rod gradually decreases from one end to the other....

Let the length of the rod be $L$.

When the rod is pivoted at one end and a horizontal force $F$ is applied at the free end perpendicular to the rod, the torque about the pivot is given by $\tau = F \times L$.

This torque depends only on the magnitude of the force and the length of the rod. It is independent of the mass distribution or which end is chosen as the pivot.

The moment of inertia $I$ of the rod depends on the mass distribution. Since the density gradually decreases from one end to the other, the moment of inertia about the heavier end will be different from the moment of inertia about the lighter end.

Angular acceleration is $\alpha = \dfrac{\tau}{I}$. Since $I$ depends on the pivot, $\alpha$ also depends on the pivot.

Angular velocity after one rotation ($\theta = 2\pi$) is $\omega = \sqrt{2\alpha(2\pi)} = \sqrt{\dfrac{4\pi \tau}{I}}$, which depends on $I$.

Angular momentum after one rotation is $L = I\omega = \sqrt{4\pi \tau I}$, which also depends on $I$.

Therefore, only the torque of the applied force does not depend on which end of the rod is pivoted.