I think I have mastered the translation process. Please allow me to assist.
Have coordinate at x' = -5/6ls.
In a 1+1 space time, let us have two inertial frames S, with unprimed coordinates x and t, and S', with primed coordinates x' and t'.
Let A be the time-like world-line containing all events in this spacetime such that $$x' = - \frac{5}{6} \, \textrm{light-seconds}$$.
What Lorentz transforms standard configuration say for elapsed time for x' to reach origin of unprimed frame at x = 0.
Assuming that the frames S and S' agree on event O as being the origin of their respective space and time coordinate systems, we call this standard configuration such that event O has the following coordinates:
$$x_O = 0, \, t_O = 0, \, x'_O = 0, \, t'_O = 0$$
Let B be the time-like world-line containing all events in this spacetime such that $$x = 0$$.
Obviously, unless the frame S and S' are not in relative motion, world-lines A and B share only one event in common. Following convention (established below) we call this event Q.
Let us call the motion of S' relative to S as v, so now we may compute the coordinates of Q in both frames:
$$x_Q = 0, \, t_Q = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}}, \, x'_Q = - \frac{5}{6} \, \textrm{light-seconds}, \, t'_Q = \frac{\frac{5}{6} \, \textrm{light-seconds}}{v}$$
t' = ( t - vx/c² )γ
t' = ( t - 0 )γ = tγ.
Applying the Lorentz transform from S to S' for Q, we see that
$$\begin{pmatrix} t'_Q \\ x'_Q \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} & \frac{-v}{c^2 \sqrt{1- \frac{v^2}{c^2}}} \\ \frac{-v}{\sqrt{1- \frac{v^2}{c^2}}} & \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} \end{pmatrix} \begin{pmatrix} t_Q \\ x_Q \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} \\ \frac{-v}{\sqrt{1- \frac{v^2}{c^2}}} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \sqrt{1 - \frac{v^2}{c^2}} \end{pmatrix} =\begin{pmatrix} \frac{\frac{5}{6} \, \textrm{light-seconds}}{v} \\ - \frac{5}{6} \, \textrm{light-seconds} \end{pmatrix}$$
So, moving clock coming toward rest origin beat not time dilated.
The poster then asserts that a clock on world-line A, which is moving at speed v according to S, is not time-dilated.
It is wrong for a number of reasons. Just because at event O (which is on world-line B, and not on world-line A) $$t_O = t'_O = 0$$ does not imply that on world-line A (the world-line of the moving clock) that $$t = t'$$. Specifically, on world-line A, t=0 and t'=0 specify two different events.
To see this parameterize the events on A with the parameter a (with units of length):
$$x_{A_a} = \frac{a}{\sqrt{1-\frac{v^2}{c^2}}}, \,
t_{A_a} = \frac{(\frac{5}{6} \, \textrm{light-seconds})(\frac{c^2-v^2}{v c^2}) + \frac{a}{v}}{\sqrt{1-\frac{v^2}{c^2}}}, \,
x'_{A_a} = - \frac{5}{6} \, \textrm{light-seconds}, \,
t'_{A_a} = \frac{(\frac{5}{6} \, \textrm{light-seconds}) + a}{v}$$
So if you wanted to answer a question about time dilation of a clock moving (according to S) on world-line A and ending at event Q, you need to choose a starting event P on A. (Once again, O is not on A.)
$$P_{t'=0} \, = \, P_{a = - \frac{5}{6} \, \textrm{light-seconds}} $$ with coordinates:
$$x = \frac{- \frac{5}{6} \, \textrm{light-seconds}}{\sqrt{1-\frac{v^2}{c^2}}}, \,
t = \frac{(\frac{5}{6} \, \textrm{light-seconds})(\frac{-v}{c^2})}{\sqrt{1-\frac{v^2}{c^2}}}, \,
x' = - \frac{5}{6} \, \textrm{light-seconds}, \,
t' = 0$$
But $$P_{t=0} \, = \, P_{a = - (1 - \frac{v^2}{c^2}) (\frac{5}{6} \, \textrm{light-seconds} )} $$ has different coordinates:
$$x = - \sqrt{1 - \frac{v^2}{c^2}} (\frac{5}{6} \, \textrm{light-seconds} ) , \,
t = 0, \,
x' = - \frac{5}{6} \, \textrm{light-seconds}, \,
t' = \frac{v (\frac{5}{6} \, \textrm{light-seconds})}{c^2}$$
So because $$P_{t'=0} \neq P_{t=0}$$ you have to be extra careful when talking about time dilation as a clock moves from P to Q, since
$$t_Q - t_P$$ depends on the choice of P.
FYI, this is likely chinglu1998 who recently appeared on various forums with contrived physical situations which he seeks to leverage into arguing that the Lorentz transform predicts absolute simultaneity or absence of time dilation for moving clocks.
Watch out for ambiguous problem statements that leave one or more endpoints of an interval ambiguously defined so that an apples and oranges comparison results or arguments that a Lorentz transformation of a uniformly moving particle may result in a frame where the particle has the same time dilation factor as the original speed. (This happens in 1+1 space time when you switch to a frame with twice the rapidity of the particle.)