In the book Explorations in Mathematical Physics (p. 262) by Don Koks, he discusses a traveler who has constant proper acceleration and therefore a Rindler horizon behind her, about which he says:
On this mysterious plane, she observes that time itself has stopped, and although she cannot see what lies beyond, she knows from our previous discussion that behind the horizon time is flowing in reverse!
This makes sense to me given that an object that is stationary in her frame has a rate of time in her frame (in Rindler coordinates) given by $\frac{d\tau}{dt} = \alpha x$ where $\alpha$ is her acceleration. Since the horizon is at $x = 0$, the rate of time is $0$ there. And since $x$ is negative on the other side of the horizon, the rate of time is negative there (although technically, the Rindler coordinate system is only defined for $x \gt 0$).
Now I realize that she won't actually see anything move backwards through time, and the same is true of time dilation—it can't be directly observed in real time; it can only be inferred after the fact. But here's the main thing I'm wondering about this reversal of time in her frame, which I'll call "time inversion" (rather than "time reversal" to avoid confusing it with the unrelated topic of time reversal symmetry/invariance): Is time inversion every bit as physically real as time dilation? If so, why do we only ever hear that time's speed is relative and not that its direction is relative as well?
If your answer is that it's not as physically real (e.g., that it's the result of coordinate changes), please read on so that I can explain why it seems to me that it must be, and please refer to this spacetime diagram:

As you've probably noticed, this basically depicts the twin paradox with an additional stationary twin/triplet (A) who is just watching the action from afar—3 light years away in fact. C travels away from B at $v = 0.6c$ $(\gamma = \frac{1}{\sqrt{1-0.6^2}} = 1.25)$ until reaching a destination $3$ ly away before returning. So he travels for a total of $\frac{2 \times 3}{0.6} = 10$ years of Earth time and $\frac{10}{1.25} = 8$ years of proper time. The blue and red lines are of course the planes of simultaneity. The blue and red numbers show C's proper time $(\tau)$. The red numbers are times when he's at rest relative to A and B, and the blue numbers are times when he's moving.
Using these numbers, you can track C's accounting of A's time, and you can see that, according to it, A's time starts at $t = 0$, then quickly drops well below $0\ ($to $-1.8)$ during C's acceleration—i.e., A's time seems to move backward—and it stays in that past for more than $2$ years of C's time while it moves forward at $80\%$ of the proper rate for the first half of the journey. Then, at the turnaround, it leaps forward by several $(7.2)$ years, and then resumes its slow advance until reaching a value well above $10\ (11.8)$. It then quickly falls backward again during C's deceleration to settle at $t = 10$. Note that for simplicity, we're assuming that all acceleration is effectively infinite (hence the backward time jumps and the forward time jump at the turnaround are instantaneous).
Here's my main argument: We know that 1) C's journey is physically $2$ years shorter in his frame than it is in A and B's frame. And we know that 2) during the inertial legs of that journey, A's time and B's time physically run at $80\%$ of the speed of C's time in C's frame. Because of these two things, we also know that 3) B's time has to physically leap forward $3.6$ years (since $10 - 80\% \times 8 = 3.6$) in C's frame during the turnaround (which can be explained either in terms of rotating planes of simultaneity or the equivalence principle of general relativity and gravitational time dilation). And the same things that require that to happen make it so that 4) A's time must physically leap forward $7.2$ years in C's frame during that same turnaround (twice as many years as B's time since A is twice as far away as B (see this page from Michael Weiss and John Baez on "distance dependence"); equivalently, A is half as deep in the apparent uniform gravitational field that we can say causes A and B to "fall" towards C). And in order for this to all be true, 5) A's time must physically move backwards by a total of $3.6$ years in C's frame at some point(s) during the journey (since $80\% \times 8 + 7.2 - 10 = 3.6$), as it seems to do. These five statements are all interdependent, so if any one of them isn't valid (i.e., the effect it describes is not physically real), the others—including (1)—are all called into question until at least one other of them is modified to compensate for it. So is (5) valid? If not, which of the other statements need to be modified to preserve (1)?
The Lorentz transformations show these backward time jumps as well. E.g., the first one happens at $t' = 0$ when $v = 0.6c$, $x = -3$, and $x' = \frac{x}{\gamma} = \frac{-3}{1.25} = -2.4$ (per the length contraction formula—if there's a better way to do this, let me know), so
\begin{align} t & = \gamma \left(t' + \frac{vx'}{c^2}\right) = 1.25 \left(0 + 0.6(-2.4)\right) \\ & = -1.8. \end{align}
(Note that since all of our units are light years (ly), years (y), and light years per year (ly/y) and since $c = 1$ ly/y, we can omit units and $c$ from our calculations, as I've done.)
So while C speeds up, he goes from knowing that A's $t = 0$ is happening currently (i.e., simultaneously with his present time) to knowing that A's $t = -1.8$ is happening currently. In other words, he believes that A's time rewinds $1.8$ years, and in his reference frame, he's correct, is he not?
While I haven't shown that these backward time jumps are due to A being behind the Rindler horizon of C, it seems safe to say that the same effect is at play. Since C's acceleration is effectively infinite, the horizon's distance from him is $\frac{c^2}{\alpha} = \frac{c^2}{\infty} = 0$, so virtually everything in the other direction is behind it.
So should we say that the relativity of time applies not just to the speed of time but to its direction as well? If time dilation is the result of one observer rotating in spacetime relative to another, causing their time to advance in a somewhat different direction, could the combination of acceleration and distance (i.e., "gravitational" potential) allow that rotation to simply continue until it has a component in the opposite direction?
Notes
This all raises questions about whether an analogous effect might exist beyond the event horizons of black holes and the cosmological horizon of the universe but those are probably topics for another day. For now, I'll just say that seeing a number of answers on this site that claim that time beyond a black hole's event horizon does not run backwards relative to ours is part of what sparked my interest in this topic (and interestingly, Koks doesn't touch this question in his chapter on black holes and GR even though he discusses horizons extensively in it).
Before writing this question, I found two similar ones (here and here) but I don't think my questions and points are addressed in either of them. I've also done some pretty deep searching elsewhere and, to my surprise, have found very little on this topic, so if you happen to know of any more resources on it, please share. All I've found so far, aside from Koks's book, is a mention in a class handout called "Space-Time Diagrams, Relativity of Simultaneity, and Relative Time Reversal" that "relative time reversal can only occur between events that cannot communicate by any physically realistic signal." While I'm sure this is true, the statement seems intended to mean that said reversal would require superluminal signals or travel and is therefore impossible. Apparently, it hadn't occurred to the author that an event horizon might fulfill the criteria.
Thanks and sorry for the long question!