Is two cars colliding at 50mph the same as one car colliding into a wall at 100 mph?

I don't think any of the other answers have made the following point clear enough, so I am going to give it a try. Both scenarios are very similar before the collision, but they differ greatly afterwards...

From a stationary reference, you see the cars driving towards each other at 50mph, but of course if you choose a reference frame moving with the first car, then the second will be headed toward it at 100 mph. How is this different from the wall scenario?

Well, from a stationary reference frame, after the crash both cars remain at rest, so the kinetic energy dissipated is $2\times \frac{1}{2}mv^2$.

From the reference frame moving with the first car, the kinetic energy before the crash is $\frac{1}{2}m(2v)^2=4\times\frac{1}{2}mv^2$, but after the crash the cars do not remain at rest, but keep moving in the direction of the second car at half the speed. So of course the kinetic energy after the crash is $2\times\frac{1}{2}mv^2$, and the total kinetic energy lost in the crash is the same as when considering a stationary reference frame.

In the car against a wall, you do have the full dissipation of a kinetic energy of $4\times\frac{1}{2}mv^2$.

Actually, assuming that the oncoming car is the same mass as yours, colliding with an oncoming car at 50 MPH is equal to colliding with an ideal immovable wall at 50 MPH. Consider this:

I'm going to set up one of two experiments. I'm either going to ram car A into car B, both of them moving 50 MPH in opposite directions, or I'm going to ram car A into a solid wall at 50 MPH. However, I'm going to put up a shroud so that you can only see car A, you will be unable to see either car B or the wall, whichever one my coin-flip tells me to use.

Because you can now only see at car A and its contents, how would you tell which experiment I'd decided to do?

I think that it makes sense to move away from the specific walls and cars and consider simply an inelastic collision on of two masses, $m_1$ and $m_2$. Otherwise we get stuck in the details.

When two bodies collide, the devastating effect in collision depends only on their relative velocity $v_1-v_2$. Kinetic energy, which has the destructive effect is equal to

$$\frac{1}{2}\frac{m_1m_2}{m_1+m_2}(v_1-v_2)^2$$

The rest of the kinetic energy is associated with the movement of the center of mass of the system. This energy in the collision does not change, and has no effect of destruction.

In given case, if faced two identical cars moving toward each other with one and the same velocity $v$ the energy of destruction is

$$\frac{1}{2}\frac{mm}{m+m}(v+v)^2=mv^2$$

Now, consider the case where a car collides with a massive barrier at speed $2v$.In this case $m_1=m$, $v_1=2v$, $m_2=\infty$, $v_2=0$

The energy of destruction:

$$\frac{1}{2}m(2v)^2=2mv^2$$

I.e. the latter case is much more dangerous.