This is a complex subject, but basically my opinion is that it is fair so long as approximately one-eighth of the boards are switched. Let me try to make this simple.
First, what is an arrow-switch? When you play a Mitchell movement, ie one where North-South [N/S] is stationary, and the boards and East-West [E/W] pairs move in opposite directions, you normally finish with two winners, one N/S and one E/W. However, this means you are never actually playing the pairs with whom you are competing: if you are N/S you play against E/W pairs but you compete against N/S pairs. If you turn some of the boards through 90 degrees so that the moving pairs play N/S and the stationary pairs E/W then every N/S pair is actually competing with some E/W pairs on every board. Then you can produce a final ranking list of all the pairs in each direction because they are all competing against each other.
So far, so good! But the problem is to find out how many boards should be arrow-switched to make it fair. Of course, instinct suggests half the boards, but that is quite wrong. Let me see if I can show you why.
Suppose you are N/S and you consider a N/S pair at the other end of the room: you switch half the boards, but so do they. In effect you seem to have finished in the opposite direction to them on about three boards in four, and in the same direction as an E/W pair on about one board in four. That's not what we are trying to do!
OK, let's try switching one board in four. That feels fairer: you switch one board in four, and the other N/S switches one board in four, on average it comes to one board in two in the other direction. Similarly you find an E/W pair is in the same direction as you on about one board in two. Goodness, perhaps we have found the solution! In fact there are some publications that recommend this, but regrettably it is wrong.
To quote Max Bavin, the EBU's Chief Tournament Director: "Some pairs don't play against each other. As compensation for this, we want them to have to compare scores more frequently with one another." To see what he means, consider how you are competing against the other pairs. If you are N/S then you are competing against the other N/S pairs on half the boards. If you get a good score, you are reducing the other N/S pairs' scores by 2 matchpoints, and so on.
Is it not the same for E/W? No, because you play one round against them, and on those boards you have a much bigger effect on them. If you were to get a top, you do not reduce the E/W pair's score by 2 matchpoints, but by a full top. The point is that there is much more competition between you and a pair you actually play against.
So to make it as fair as possible, you need to compete directly on more boards against pairs that you do not actually play, so you need to switch fewer than one board in four. Aha, you say, but how do we calculate it as one in eight? I am afraid that only a complex mathematical argument will do this last step, but if I have convinced you that arrow-switching is fair, and that it should be less than one board in four, then I have done all I can do. There is a section on my Lawspage on the web with some articles on this subject, by Max Bavin, John Probst, David Martin and one by John Manning, the expert in the field, regrettably no longer with us.
Trust me: arrow-switching is fair, and roughly one board in eight should be switched!