Ok, that is true:
1.a4 h5 2.a5 h4 3.a6 h3 4.axb7 a5 5.d4 a4 6.Bg5 a3
7.e3 a2 8.Qf3 axb1=B 9.Qf6 exf6 10.Be2 Bd6 11.Bg4 Bf4
12.Nf3 d6 13.0-0 hxg2 14.h4 Bf5 15.h5 Kd7 16.h6 Kc6
17.h7 Nd7 18.hxg8=R Rh3 19.Rh8 Rg3 20.Rh1 Qh8
21.b8=R Qh5 22.Rh8 Rg8 23.Ra8 Ba2 24.Ra1
Almost a cyclic permutation of 4 wRs, since wRh8 came from b8, not from a8. I think that in the retrograde analysis corner problem the cycle was perfect. Maybe I'll find that problem later.