Y is $f_t$-related to Y implies $f_t(g_s(p)=g_s(f_t(p))$ for all p$\in M$?

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I saw this claim, but stuck on this for a long time: $X,Y\in \Gamma(TM$) are two complete vector fields with flows {$f_t$},{$g_s$}, then Y is $f_t$-related to Y implies $f_t(g_s(p)=g_s(f_t(p))$ for all p$\in M$.