If $x(t)$ be a known trajectory does the $x(-t)$ represent the retracing trajectory? Assertion
If there is time-reversal invariance, Newton's law (for a system described by one generalized coordinate $q$) $$m\frac{d^2}{dt^2}q(t)=F\Big(q(t)\Big)\tag{a}$$ implies that if $q(t)$ is a solution, $q(-t)$ is also a solution i.e., $$m\frac{d^2}{dt^2}q(-t)=F\Big(q(-t)\Big)\tag{a1}$$ The operational implementation of time-reversal $t\to -t$ requires doing the following: $$q\to q,~~\text{and}~~\dot{q}\to -\dot{q}\tag{b}$$ to the instantaneous values of $q$ and $\dot{q}$ by which the system is made to retrace the path.
A concrete example
Suppose a harmonic oscillator starts from right extreme position A with initial conditions $x=a$ and $\dot{x}=0$ at $t=0$. Therefore, its trajectory is given by $$x(t)=a\cos\omega t.\tag{1}$$ It reaches the mean position B at time $t=T/4$ at which $x=0$ and $\dot{x}=-a\omega$. Now the prescrition of the time-reversal (Eq. (b)) requires that at $t=T/4$, we set $x=0$ and $\dot{x}=+a\omega$, and verify whether the trajectory is retraced. It's easy to check that with these as the initial conditions, the solution becomes $$x(t)=-a\cos\omega t\tag{2}$$ which indeed represents the retracing path BA traversed from B to A i.e., the trajectory (2) is opposite to the trajectory described by (1).
$\bullet$ However, note that the retracing trajectory (2) cannot be obtained from trajectory (1) by simply sending $t\to -t$, and therefore, I have the following question.
Question
If $x(t)$ represents a continuous trajectory AB traversed in the direction from A to B, which trajectory should $x(-t)$ be identified with? From my example, it appears that $x(-t)$ is not the retracing trajectory.
 A: This is just a semantics question.


*

*$x(-t)$ is the time reversed trajectory of $x(t)$. Physically, for $t > 0$, you can imagine this trajectory as that of a particle that starts moving at negative $t$, then is time reversed (intuitively, 'hits a mirror') at $t = 0$.

*the trajectory that "reverses the path" of $x(t)$ by time reversal at time $t_0$ is $x(-t + 2 t_0)$. Note that whenever we talk about "time reversal" alone, we always mean time reversal at time $t = 0$.

*if time reversal symmetry holds and $x(t)$ is a solution, $x(-t)$ is a solution

*if time reversal symmetry and time translational symmetry hold, and $x(t)$ is a solution, then the "reversed path" $x(-t + 2 t_0)$ is a solution


Most of the time, when time reversal symmetry is present, time translational symmetry is also present, so we don't bother to distinguish these two concepts. 
On the other hand, consider a particle in the electric field $E(t) = E_0 (t /t_0)^2$. This electric field obeys time reversal symmetry (about the time $t = 0$, as by convention) but it breaks time translational symmetry. Therefore, if $x(t)$ is a solution, so is $x(-t)$, but $x(-t + 2 t_0)$ is not, as you can check.
A: I don't know the physics, but I can speak a little bit about the math. This is an exercise in function transformations: Vertical shifts, Vertical scaling/reflecting, Horizontal shifts, and Horizontal scaling/reflecting.
$f(x) + 10$ would be a vertical shift, $-2f(x)$ would be a vertical stretch + a vertical reflection or vice versa, and $f(-x)$ is a horizontal (left/right) reflection of the graph across the y-axis.
So imagine a graph of $x(t)$. You start your pen at negative infinity and draw some curve all the way to positive infinity. Now $x(-t)$ is mathematically speaking, just the left/right reflection across the y-axis. Therefore all outputs that happened after $t = 0$ suddenly get swapped to the negative side of the time axis. That is, everything that happens last suddenly happens first. And everything that happened first (on the negative side of $t$) now happens last after you do the reflection. While $x(t)$ traces out some path from the beginning of time to the end of time. $x(-t)$ would trace out the exact same path, but in reverse (it plots out positions of $x(t)$ in reverse).
So now imagine starting your pen at negative infinity. You draw some $x(t)$ curve all the way up to $t = 0$ and stop. If you want the reflected graph on the other side of $t = 0$ just make $x(-t)$ on the other side. Therefore as you start moving the pen again from $t = 0$ onward to infinity, you just retrace everything you did in reverse. Now let's look at another answer to this question. Say you start the pen at negative infinity and draw some $x(t)$ curve all the way to $t = t_0$. And let's assume $t_0$ is some positive number (we don't have to, as it's a variable. But let's do it anyways). You stop your pen at $t_0$. From $t_0$ onward, you want the backtrack. Therefore, you need the reflected graph of $x(t)$ across the line $t = t_0$. So you say, hmmmm. What if I make the reflected graph across the y-axis $x(-t)$? But then you realize you need to push the graph to the right by $2t_0$. This is a combination of function transformation. Combining 2 horizontal transformations is a little harder than 2 vertical transformation. But nonetheless, the graph on the right side of $t = t_0$ will be $x(-(t - 2t_0))$ or $x(-t + 2t_0)$, which is the knzhou's response.  
