In what proportions do length and time adjust to preserve the speed of light? Suppose my friend is moving relative to me in the direction of motion of a light pulse with a speed $\frac{c}{2}$. We both have identical meter scales and clocks.Then, the universe can preserve the speed of light with respect to him by making one of these adjustments:


*

*His scale becomes half of my scale and his clock continues ticking with the same rate.

*His clock ticks slows down to half the speed of my clock, and his scale remains unchanged.

*His clock slows down to $\frac{3}{4}$ of its original speed, and his scale contracts to $\frac{2}{3}$ of its original length.
etc....
So, length and time can adjust in multiple proportions in order to preserve the speed of light. How is it decided that which parameter is adjusted to what extent?
 A: From your point of view, your friend appears to be a) aging more slowly then when they are at rest with respect to you, and b) "flattened" in the direction of their movement. More specifically, these quantities are "off" by a factor of $\gamma = 1/\sqrt{1-v^2/c^2}$, where $v$ is your friend's velocity. These are always the adjustments made to "preserve" the speed of light. In your example, $\gamma = 2/\sqrt3 \approx1.15$. So, the period of your friend's clock is $1.15$ times as long as yours, and their scale is $1.15$ times shorter than yours. Indeed, this is the only possible way that $c$ can be the same in all refernce frames. 
EDIT- Here, I'll derive the formulas for time dilation and length contraction mentioned above. 
Imagine that your friend is in a train cart of height $L$ moving at velocity $v$. Say the cart has mirrors on the ceiling and floor, and that he has a clock which functions by ticking every time a light beam (which is repeatedly traveling from the floor to the ceiling and back) hits the floor. From his view, the period of ticks is $\Delta t = 2L/c$. However, from my point of view, say the cart travels a distance $x$ in between each tick. Then the light travels a distance $\sqrt{x^2+4L^2}$ between each tick, and so the period of ticks is (noting that $x = v\Delta t'$)
$$\Delta t' = \frac{\sqrt{x^2+4L^2}}{c} = \frac{\sqrt{v^2\Delta t'^2+4L^2}}{c}$$
Solving for $\Delta t'$ gives us
$$\Delta t' = \frac{2L}{\sqrt{c^2-v^2}} = \frac{2L/c}{\sqrt{1-v^2/c^2}} = \gamma \Delta t$$
which gives us the formula I described above. As for length contraction, suppose a particle decays in time $t$, where this time is measured from its own reference frame. Further, suppose it's traveling at a velocity $v$ towards an observer. As we saw above, to the observer, the particle will take $\gamma t$ to decay. Thus, it travels $L' = v\gamma t$ in its lifetime, from the point of view of the observer. However, in the particle's reference frame, it travels a distance $L = vt$. Thus, we have $L' = \gamma L$ (this is to say, the environment looks "shorter" to the particle in the direction of its motion). 
