I'm pretty sure I'm misinterpreting something, but my reasoning for why it is not is that when you reverse time, the trajectory in which $x$ follows, namely $x(t)$, changes direction, so that at any given nonzero time {$t_0 | t_0 \ne 0$}, $$x(t_0) \ne x(-t_0)$$ in general.
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2$\begingroup$ A very absurd example: London is still London if the time runs backwards. $\endgroup$– orionCommented Jul 21, 2015 at 8:08
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That argument is correct, but it applies to a trajectory, which is a function mapping times to position vectors. A single position vector itself, something like $(-5,2,3)$, is not associated with time in any way, and when you reverse time, it doesn't change the point that is labeled by those coordinates.
Maybe this simplified example will help: consider a function $f(t) = t^2 + 2t$. The function itself is not time reversal invariant, because $f(t) \neq f(-t)$. But each possible value of the function is invariant. $5$ is not affected by taking $t\to -t$.
