The pulses reflect from the mirrors at the same time; events on the blue are simultaneous.
But light waves travel through empty space, so there is no way to tell which apparatus is moving. Thus, the reflections also happen at the same time in the moving frame, and the $x'$ axis is parallel to the new blue line!
Both observers must agree on the area of any figure in the spacetime diagram - any difference would distinguish between relative motion to the left and right.
Therefore the clocks will read the same at the end of the experiments if and only if the areas of the two gold light path diamonds are equal.
In Euclidean geometry, the invariance of area under rotation is also how you compare rotated and unrotated coordinates.
In Euclidean geometry, $(-b,a)$ would have been orthogonal to $(a,b)$.
The light blue figure is a spacetime square. Adjacent edges are orthogonal in the spacetime sense. The length of the spacelike edges is the distance light travels in the duration of the timelike edges, so the diagonals are light paths.
The area of the light blue spacetime square is $a^2-b^2$ (the same as in Euclidean geometry).
In primed coordinates, $(b,a) \mapsto (0,c\tau)$ and $(a,b) \mapsto (c\tau,0),$ so the area of the light blue square is simply $c^2\tau^2 = a^2-b^2.$ $\tau$ is the proper time between $(0,0)$ and $(b,a).$
The tangent to the hyperbola at P is spacetime orthogonal to world line OP, with slope reciprocal to the slope of OP.
The ratio between $ct$ at point P and at point V is $\gamma=1/\sqrt{1-v^2/c^2}$. The trip takes time $\gamma\tau$ for a non-traveler.
You cannot travel “faster than light” only from the point of view of a stationary observer. A traveler can go any distance in any given time $\tau$.
The tangent to the hyperbola at P is spacetime orthogonal to OP, with slope reciprocal to the slope of OP.
This timelike hyperbola is the world line of an object moving with constant acceleration $c^2/a$.
One Earth gravity acceleration has a distance $a$ of roughly one light year.