$$x_{n+1}=x_{n}\mu \left( 1-x_{n}\right)$$ this has a 'STABLE' period 1 orbit with 2 fixed points upto a certain parameter value,its a fact.
$$x_{n+2}=x_{n+1}\mu \left( 1-x_{n+1}\right)$$
For the next iteration,it would have a 'STABLE' period 2 orbit with 4 fixed points for a certain range of parameter values
Calculation of fixed points for period 2 orbit equation yields:
$$x_{n+2} = \mu^2 x_n(1-x_n)(1-\mu x_n(1-x_n))$$
So,to gather the fixed points,we set:
$$x_{n+2} =x_n$$
implies
$$x_n^* = \mu^2 x_n^*(1-x_n^*)(1-\mu x_n^*(1-x_n^*))$$,where $$x_n^*$$ are the fixed points
In order to solve the above expression, my book assumes that 2 of the roots are 'already known' i.e they are the fixed points of the period 1 orbit and hence continues to find the other 2 roots,thereby obtaining all 4 roots of the period 2.
Why would fixed points of the period 1 orbit be the same as the fixed points of the period 2 orbit?I'm trying to visualize this in terms of Poincare' maps but it doesn't make sense why the above assumption is true.
(Just for reference,I add that the fixed points of the period 1 orbit are 0 ,(1-1/mu))