In the following question I am trying to find the fixed points, stability, and then trying to see what it converges to.
To find the stability of this we must find $|f'(x^*)|$ to see whether it is $>1$ which would make it unstable or $<1$ which would make it stable.
$x_{n+1}=lnx_n$
So setting $f(x)=lnx$ and $f(x)=x$ to find the fixed point then $x=lnx$
But how do I find the value of this $x$ that makes $x=lnx$ work. You cant use $x=1$ because $1\ne 0$. I am not trying to use a graphical method either, trying to work it out numerically.
Once I have the fixed points, I could then use $|f'(x^*)|$ to find out whether $|f'(x^*)|<1$ to make it stable or whether $|f'(x^*)|>1$ which makes it unstable.
Are there any intersection points?
Let us rewrite the equation $x = \ln x$ satisfied by the stationnary points as $$ {-1} = {-\ln x}\, \text{e}^{-\ln x} \, , $$ which is of the form $X \text{e}^{X}= Y$ with $X={-\ln x}$ and $Y={-1}$. The solutions are $X=W (Y)$, where $W$ is the Lambert W-function. Hence, $$ x = \text{e}^{-W (-1)} = -W (-1) \, . $$ This value is not real, since the natural logarithm $x\mapsto \ln x$ and the identity $x\mapsto x$ don't cross on the real line.
When studying the present dynamical system, one can observe that $x_{n+1} = \ln x_n \leq x_n -1$. The sequence $x_{n+1}= x_n -1$ is strictly decreasing with no limit, and so does $x_{n+1} = \ln x_n$ (until $x_n \leq 0$ for some $n$, where $x_{n+1}$ becomes complex).