Solved with Topdecher. Under inversion with center $S$ and radius $SA_1$ point $A_i$ is fixed and point $X_i$ is mapped onto midpoint $M_i$ of $A_iA_{i+1}.$ It's suffice to prove that all circles $(M_iA_{i+1}M_{i+1})$ concur. Indeed, if $O$ is a projection of $S$ onto base of pyramid, then all triangles $OSA_i$ are congruent. Hence $O$ is a circumcenter of $A_1A_2...A_n,$ which clearly is desired concurrency point.