Attempt to a partial answer.
I expect this to be impossible, but not by principal means (i.e. a real proof will be very hard and contain many cases), but the limitations of the pieces and the board. To show this, here is a "mild" fairy solution using nothing than a large board and some fancy line figures.
Start with Kc4 Rd5 Rd4 Pd6 Pe5 Pe4 Pd3 - Ka4 Ra6 Ra3 Be6 Be3 Pd7. The intended perpetual is 1.Kc5+ Ka5 2.Ka4+ Kc4. You now "merely" need white girafferiders (a line figure on 1,4-lines) on i2 and i7 to pin the bishops, and four more fancy riders to pin the girafferiders in both positions of the white king.
This setup also shows where the problem is: Why shall White's checking piece not move to the last position of the black king in the next White move? Either because White is in check (but then indeed we run into the problem of a mutual perpetual check, as outlined in my comments) or because it is pinned. But then, why shall the pinner not capture instead? It must be pinned too. And that pinner is White again and must not move either, i.e. must be pinned, leading to some setup like Kh1 Rf3 Bb1 - Kf5 Be4 Ra1. There are only very few of these setups, and worse, since it must be an infinite cycle on a finite board, all moves must ultimately be "taken back", so the Rf3 here must be unpinned somewhen again. How?
