Why not just average the 5 CDFs? The average (which is 1/5 of the sum) is a monotone function to [0,1]. This would also be the mathematically reasonable thing to do.
More specifically, a CDF is F(x) = Prob(X < x), and the average CDF would be Favg(x) = Prob(choose X from X_1, X_2… X_5 with equal prob AND X < x)
Or another interpretation is, Favg is the CDF of a variable X obtained by choosing one of the CDFs uniformly and sampling a value according to it.