Not in this frame no.
No its that only in the frame where the box is at rest does the light reach the edges of the box at the same time but in any frame where it is moving they don't because "same time" is not frame invariant not even if you write it in all caps.
The correct way to understand this is to use the Lorentz transforms but I'm gonna need the inverse transforms so I'll state those
\[\begin{eqnarray*}
t&=&\gamma\left(t'+\frac{v}{c^2}x'\right)\\
x&=&\gamma\left(x'+vt'\right)\\
y&=&y'\\
z&=&z'
\end{eqnarray*}\]
And \(\gamma=\left(1-\frac{v^2}{c^2}\right)^{-1/2}\). Now I can work in the primed frame where the box is at rest and if I choose the origin to be the middle of the box when the light is turned on then it's easy to write the coordinates of the receiver events so your Z receiver receives light at \(x'=0,\ z'=0.5,\ t'=0.5\) if I measure distance in light seconds and time in seconds and I ignore \(y'\) because it's always zero for everything and your X receiver receives light at \(x'=0.5,\ z'=0,\ t'=0.5\). You chose \(v=0.638971c\) which makes \(\gamma=1.3\) and it's easy to feed that through the inverse transforms to get that in the frame where the box is moving your Z receiver receives light at \(x=0.4153,\ z=0.5,\ t=0.65\) and your X receiver receives light at \(x=1.0653,\ z=0,\ t=1.0653\) and you can see the times aren't the same and
relativity has no problem with this it's only you who gets all confused by it. You can easily see that light moved 0.5 light seconds in 0.5 seconds in the box frame and if you calculate \(\sqrt{0.4153^2+0.5^2}=0.65\) you can see that light moved 0.65 light seconds in 0.65 seconds and 1.0653 light seconds in 1.0653 seconds in the unprimed frame where the box is moving so it's always travelling at one light second per second in both frames.
If you go and compare my figures to the ones in your picture you'll see that I've got the same answer as you did for when the light reaches your Z receiver but a different answer for when it reaches the X receiver and that's because you forgot about length contraction so you drew a box that's square in a frame of reference where it's contracted in its direction of motion by 1/1.3 so would only be 0.769 light seconds long if it was a cube in its rest frame. If you want to work with a box that's 1 light second long in the frame where it's moving at 0.638971c then you need to make it 1.3 light seconds long in the x' direction and 1 light second in the z' direction in its rest frame so it isn't a cube there and in that case the X receiver receives light at \(x'=0.65,\ z'=0,\ t'=0.65\) which inverse transforms to \(x=1.1385,\ z=0,\ t=1.1385\) which is what you got.
So yeah relativity has no problem with light arriving simultaneously in some frames and not others. I realise that you've probably missed that because it's only stated in every single work on relativity ever even
Einstein's first paper where he says "the observers moving with the moving rod, thus would not find the clocks synchronous, though the observers in the stationary system would declare the clocks to be synchronous" in the last but one paragraph of
§2 so you might have missed it if you'd never even looked at a serious source. And I know you have trouble parsing simple sentences because that's what got this conversation started so before you get your knickers in a twist over Einstein talking about the stationary system he explicitly states that it's an arbitrarily chosen system that he's decided to call stationary for convenience at the top of
§1.
