On the effect of symmetry requirement for rendezvous on the complete graph

We consider a classic rendezvous game where two players try to meet each other on a set of $n$ locations. In each round, every player visits one of the locations and the game finishes when the players meet at the same location. The goal is to devise strategies for both players that minimize the expected waiting time till the rendezvous. In the asymmetric case, when the strategies of the players may differ, it is known that the optimum expected waiting time of $\frac{n+1}{2}$ is achieved by the wait-for-mommy pair of strategies, where one of the players stays at one location for $n$ rounds, while the other player searches through all the $n$ locations in a random order. However, if we insist that the players are symmetric --- they are expected to follow the same strategy --- then the best known strategy, proposed by Anderson and Weber, achieves an asymptotic expected waiting time of $0.829 n$. We show that the symmetry requirement indeed implies that the expected waiting time needs to be asymptotically larger than in the asymmetric case. Precisely, we prove that for every $n\geqslant 2$, if the players need to employ the same strategy, then the expected waiting time is at least $\frac{n+1}{2}+\varepsilon n$, where $\varepsilon=2^{-36}$.