What is the air pressure on a magic wall that reflects all incident molecules perpendicularly to itself? I am a total laywoman in physics and humbly hope that physics experts of this SE can help me find the answer tо my question. A friend of mine and I had an argument about that question, and I am very much curious to find out who was right. Despite my best efforts, I have been unable to find the answer on the Internet.
Here is the question. Let's suppose we have a magic sheet that reflects all incident molecules perpendicularly to itself, but preserves their speed. That is, the velocity of each molecule right after the hit is directed strictly perpendicular to the surface no matter the incident angle and has precisely the same magnitude as before the hit. The sheet is 1 m x 1 m and is absolutely firm, and we take it outdoors. What will be the air pressure on that sheet there and why?

I humbly hope to read enlightening responses by physics experts. Since I am a student who studies something completely unrelated to physics and only vaguely remembers the basics of physics taught at school, I would be especially grateful for answers written in layman's terms.

UPDATE: S. McGrew and Wolphram jonny say that the pressure on the magic wall is different from the atmospheric one, and this coincides with what my friend told me during the argument. What I cannot understand is the following. As I vaguely remember, the air pressure is basically the weight of the entire atmosphere. That is, if we put our magic sheet horizontally, then the air pressure on it will be the weight of the entire air column above it, divided by the area of the sheet. I do not see how the exact mechanism of the reflection of molecules by the sheet can change that. So my guess was that the pressure on the magic wall will be equal to the atmospheric pressure. Exactly where do I make a mistake in my reasoning?

UPDATE 2: I am still confused as to exactly what is wrong with my argument. Let's place our magic wall horizontally and consider the entire air column above the wall. The forces acting on that air column are (i) the force of gravity and (ii) the force from the magic wall. Since it is a steady state, these two forces must balance each other. If they don't, it isn't a steady state, in the first place. According to Newton's third law, the force exerted on the air column by the magic wall is equal to the force exerted by the air column on the magic wall. So the pressure on the magic wall is equal to M g / S, where M is the mass of the air column and S is the area of the magic wall. And M can't be noticeably different from the mass in the case of a normal wall. So the pressure on the magic wall is equal to 10^5 Pa, the atmospheric pressure. Exactly where is my mistake? 
Furthermore, I cannot understand the reasoning behind the answers and comments saying that the pressure on the magic wall will be larger than the atmospheric pressure by a factor of 1.5 or so. The authors of those answers and comments seem to be focused on the change of a molecule's momentum, but who said the air density at the magic wall (i.e., within the collision length from it) will be exactly the same as at a normal wall? You need to take the difference into account, don't you? Furthermore, the magic wall will obviously produce a very peculiar velocity distribution near it, an anisotropic one. The stream of molecules departing the magic wall strictly perpendicular to it will surely mess things in the boundary layer near the wall and, in particular, affect the velocity distribution of incoming molecules by colliding with them. I can't accept the momentum argument if you ignore that elephant in the room.
 A: The pressure experienced by the magic plate would be more than a normal plate. The exact factor is minutiae. 
skip to jump below if you already have an understanding of how pressure comes about

As I vaguely remember, the air pressure is basically the weight of the entire atmosphere. ...Exactly where do I make a mistake in my reasoning?

Think about how pressure actually is generated. As you said its the weight of the atmosphere. Weight is the layman's term for the force of gravity exerted on a body. So if a solid block were kept on a normal plate, it would exert a pressure exactly given by what you said--the weight per unit area.    
For fluids (air,water etc), there is a slightly round-about way of thinking about this. They comprise of tiny particles (molecules/atoms) that are moving around in space with random speeds in random directions. When these hit any surface they get deflected.
To deflect something moving, we need to apply a force. You must have heard about this famous law that every action has an equal and opposite reaction: here reaction and action also mean force. So as a plate deflects a particle, in so doing the particle exerts and an equal and opposite force on the plate. This force is the reason for the pressure. (To calculate the average pressure of a fluid, we use statistics and some simple math to add up the force of all particles hitting the plate)
Note that as long as there is motion in the gas (this is characterized by how warm a gas is), the molecules will hit any surface submerged in it. This doesn't need the force of gravity to be present. For e.g. a box filled with gas at $1$ atm pressure placed far away from earth would expereince the same pressure on its inner walls as  that by a plate on earth's surface. 
So you see, the pressure experienced by a fluid on a surface comes about from the force exerted by the constituents which in turn comes from the way they get deflected.
For normal surfaces, we assume that only a force perpendicular to the surface can be exerted when anything hits it. This is why usually things get reflected when deflected. (you already seem to know that we also think of the collision to be lossless in energy). This is the usual way things experience pressure.
jump:
For the magic wall that you have described, there is a little bit more pressure: not only does it need to exert a perpendicular force to deflect a particle perpendicularly, which is the same as with a normal wall, it also needs to transfer any horizontal(parallel to plate) motion the particle had before, to vertical motion to keep the magnitude of the momentum same. This requires more force than before.

You mentioned that the weight of the atmosphere above a plate can be used to calculate the pressure. You correctly argued that the weight of a column of atmosphere doesn't depend on how that column interacts with the wall. However, as seen above, the pressure does.
Think about it this way: if you introduced your magic plate into a finite column of ordinary fluid, after a long enough time, all of the fluid would start moving perpendicularly away from the plate. This is same as if the plate had a rocket engine strapped to it exhausting the fluid thus pushing the plate. For a normal plate, this doesn't happen as the particles are reflected randomly. 
A: According to the kinetic theory of gases, the pressure in the wall results from the collision of molecules. If the area of the wall is defined by the OP to $1m^2$, we can talk about forces instead of pressure as $F = pA$.
The force in the wall in any given time $\Delta t$ is:
$F = n \frac{\Delta P_{av}}{\Delta t}$
where $\Delta P_{av} =$  average change in y-momentum of each molecule. I am considering $y$ the direction perpendicular to the wall. $n$ is the number of molecules hitting the wall in that time interval.
If the approaching molecules were free to come, independent of the reflected ones, $\Delta P_{av}$ was bigger for the magic wall, what was shown in the other answers.
But it seems to me that near the wall, a flow of reflected molecules will interfere with the approaching ones. And molecules coming from the sides, almost parallel to the wall don't have such interference. That interference decreases $\Delta P_{av}$ because exactly the approaching molecules perperdicular to the wall have the biggest change in momentum.
On the other hand it is not clear if $n$ changes when a simple wall is replaced by a magic one. 
That effects acts in the opposite direction, and may partly or total compensates the pressure increase. 
But I don't know how to model it to get a complete picture of the outcome.
A: If I understand the question correctly, your magic mirror would reflect every molecule in the same direction, regardless of what direction the molecule was moving in just before it hit the mirror.  And, the velocity of the molecule would be unchanged.
I think you want a qualitative answer, not a mathematical or numerical answer.  I'll try:
Imagine a sphere centered on a point on your magic mirror.  The net pressure on a wall at that point without that magic mirror would be proportional to the sum (integral) of *the normal (perpendicular) component of all unit vectors pointing toward the center of the sphere from the surface of the sphere.  In other words, each vector would be multiplied by the cosine of its angle away from perpendicular.
With your magic mirror, though, each vector would simply be added to the others, without the cosine factor reducing its value.  So, the resulting pressure would be greater with your magic mirror.  You can obtain the precise ratio by doing the integrals.  My math skills are a bit untrustworthy, but I think the result is that the pressure is doubled by your magic mirror.
A: The magic sheet would work roughly like a fan. Once you switch a fan on, the pressure on its surfaces increases. Your magic sheet in fact would produce a noticeable air current, wind.
