My intuition: In a system without air resistance, I would expect an arrow shot at an angle with its head pointing upwards to follow a ballistic trajectory without rotating around the horizon - because gravity can't induce torque on the body. So the arrow will impact the ground on its tail - not its head. Is this correct?
If yes, what causes the same arrow to impact the ground on its head when there is air resistance?
Your intuition in your first paragraph seems correct.
As to your question, if you think about the design of an arrow, the back of the arrow is much lighter (therefore more susceptible to being pushed around by air resistance) and designed such that the shaft of the arrow lines up with the tangent line of the arc drawn by the head of the arrow as it flies through the air.
If you imagine freezing time at some point while the arrow is on its trajectory, the location of the tail that equalizes the air resistance forces on it is when it is pointing directly backwards from the head of the arrow. Then note that the "backwards" direction changes as the arrow flies through the air under gravity, with the tail of the arrow following this change in direction.
When the head of the arrow is on its way down towards the ground, the "backwards" direction points slightly up, depending on the angle at which the arrow was fired, hence the head lands before the tail.
Another cause of the tip-down attitude is that when fired, the head end of the arrow becomes unsupported before the tail end does and hence begins to fall before the tail end does, naturally tipping the head down.
