Would there be any effects on an object experiencing "inverse" time dilation? Could "inverse" time dilation exist?

In the image below you will see a black hole and an object that is orbiting around it with a spin on an axis perpendicular to the orbit. The object is experiencing both gravitational and special relativistic time dilating effects.

To get the total time dilating effect you have to multiply the two together as seen in this post.TD Calc

I highlighted the high and low points on the orbiting object to help describe this next part.

At the low point the object is spinning such that it is not moving relative to the black holes gravitational field. This is similar to a train wheel not moving relative to the tracks at the point that touches the tracks. The high point on the other hand is moving very fast and therefore is experiencing a high time dilating effect relative to the black hole due to special relativity.

Is there a scenario that exists in which the high point is experiencing a greater total time dilation effect than the low point and if so how would this effect the orbit? Would it effect the orbit?

The reason I titled this using the words "inverse time dilation" is because time dilating effects are typically greater as you get further into a gravitational field and this would "invert" that.  We are observing this particle from very far away ("from infinity"), but we use the local static frame to measure the particle velocity $v$. The computation of time-dilation is then analogous to the computation in the question you have linked. We are also not asking how does it come about that the particle is moving on that trajectory and I will make the computation very general so that it is easy to modify for any other similar trajectory.
At the peak, the point at which the particle is farthest from the black hole, it experiences the smallest gravitational time-dilation. However, it is also fastest with respect to the local static frame and will thus have more time dilation coming from this source. Let's denote the largest Schwarzschild radius the particle is at as $r_{peak}$ and the velocity with respect to the static frame at that point as $v_{peak}$. The number of seconds $N_{peak}$ we see passing on our watch at infinity per every second elapsed in terms of proper time of the particle is then called the time-lapse factor and it reads $$N_{peak} = \sqrt{\frac{1 - 2G M/(c^2 r_{peak})}{1 - v_{peak}^2/c^2}}$$
When, on the other hand, the particle is at the bottom of the epicycle, the point closest to the black hole, it experiences more time-dilation coming from being deeper in the gravitational well, but it is static with respect to the static frame so it feels no time dilation coming from the kinematics. We can then write the time-lapse factor as $$N_{bottom} = \sqrt{1 - 2G M/(c^2 r_{bottom})}$$ Now it is simple algebra to get that time-lapse at the peak of the epicycle will be larger than at the bottom only if the peak velocity is $$\frac{v_{peak}}{c} > \sqrt{ 1 - \frac{1 - 2G M/(c^2 r_{peak})}{1 - 2G M/(c^2 r_{bottom})}}$$ You can then see that when we aren't very close to the black hole horizon ($r_H = 2GM/c^2$) and when the difference between the peak and bottom radii of the epicycle $r_{peak}$ and $r_{bottom}$ isn't too large, we can always find a velocity $v_{peak}$ such that the particle experiences more time dilation in the peak than in the bottom.