Hmm, our theoretical distribution has lambda 1. And we're making the assumption that our data x1,...xn is also exponentially distributed but with unknown lambda. One way we can estimate the parameter of the distribution of our data is to get the sample mean which is equivalent to 1/lambda and we can retrieve our estimate of lambda from there. Now we have to rescale it by applying lambda to x which makes it exp(1) to match our theoretical distribution thus allowing us to compare the plotted points on y=x.
Okay, I guess a numerical example helps to shed some light into this. Imagine we have \((x_1,...,x_4)=(1,2,3,4)\) i.i.d. \( Exp(\lambda)\). So \(\bar{X}=2.5 \) which means \( {\lambda}=0.4\) so multiplying \(\lambda=0.4\) to our data we get \((\lambda x_1,..., \lambda x_4)=(0.4,0.8,1.2,1.6)\) which now has mean 1 or likewise \(\lambda = 1\). Hence, this allows us to compare it with our theoretical distribution of Exp(1)