First of all: your solution is correct, the friction is one third of what it would have been for a "non-round", as you call it, body.
How does friction know. I like such questions, but I don't see a convincing simple reasoning right now (but there probably is one), so I will use a little of computation. I hope it's different (less computating^^) than your solution. And I hope you can follow my wishy-washy treatment, it should convey the idea to someone who can compute it by himself...:
You probably accept, that friction wants to achieve $v=\omega r$, right? Otherwise there would be slipping. Now a special property of the system comes in: the moment of inertia of a cylinder has the factor $1/2$. This means, that the angular momentum is $L=\frac 12 pr$ (it would have been equal to $pr$ for a point) if $v=\omega r$.
If this is valid all the time, the derivatives have to be equal, too. So for the total force and the total torque you also get the ratio $\tau = \frac 12 Fr$. The torque (around the center) comes only from friction (times $r$ of course), whereas the net force is the downhill gravity component $\frac{mg}2$ minus the friction. So if you want them to have the ratio of 1:2 you have to adjust friction to one third of the downhill component. If it were different, then velocity and angular velocity would change not in the right proportion to keep equal. Which means slipping.
... as another viewpoint:
You can describe the combined translation and rotation around the center by just one rotation at each instant. It rotates around the boundary point (which moves, but at each instant this description is true).
Now you can do analogous arguments with respect to this point, only with a moment of inertia of $\frac 32$
So the point is: $v=\omega r$ in this case is the equivalent of not moving in the non-rotating case.
By the way you can at once compute the necessary friction if the body is not a cylinder but a sphere: then torque and force have to be at the ratio 2:5, you need a friction force of $\frac 27$ of the gravity component $\frac {mg}2$
You are mixing up concepts.
The kinetic friction of a car has nothing to do with $F_N$.
The concept of friction as taught in school tells you, that friction is independent of velocity. This is clearly not realistic for a car! The friction - let's rather call it "air resistance", for this is the main part - has a very complicated dependence on the velocity - but it usually rises if you drive faster :)
You can keep at constant speed if all forces on you balance: the air resistance, that tries to stop you and the force that pushes you forward.
What is this force? Again friction. If there were no friction, the tires would move backwards on the road. So this is the force pushing you. For this friction the concept of static friction is suited in a good approximation.
So to you side-question: yes, for the concept of friction you learned in school, it will remain the same if you fix $F_N$ and $\mu$ - in particular independent of velociry. But this is not how reality works.
Best Answer
Sure.
If the wheel were massless, that would be true. 100% of the torque from the engine would be "delivered" to the ground. But the wheel is not massless. Some of the torque is instead going into accelerating the wheel.
Using the sum of torques formula:
$$\tau_{net} = I\alpha$$
If the wheels rotation is accelerating, then there must be a non-zero net torque. Therefore the torque from the axle and the torque from the ground cannot be equal.
There is a force pair where the force from the tire on the ground is exactly equal to the force from the ground on the tire. But when examined as a torque, these will be less than the torque from the engine.
One way to think about this is the linear example of pushing two blocks forward. You push on A, and A pushes on B. If A is massless, B is pushed with an equal magnitude force.
But the more mass A has, the smaller the force A pushes on B. In your car scenario, the larger the moment of inertia of the wheel, the smaller the torque that the ground creates.
No. A massless wheel can roll or accelerate. It just cannot have any net torque applied. Any torque placed on it must be countered by an equal torque elsewhere. It's an ideal limit.
You have to solve simultaneous equations for the system. You know the moment of inertia for the wheel, and you know the mass of the car.
The torque by the ground on the wheel will be an amount between zero and the engine torque, and that allows the wheel acceleration to match the vehicle acceleration.
If the torque is too high, the vehicle moves faster than the wheel turns. If the torque is too low, the wheel spins faster than the vehicle moves.
$$\alpha = \frac{a}{t}$$ $$F_{road} = ma$$ $$\tau_{engine} - (F_{road} \times r) = I\alpha$$