A plot of the function $y_2=b^{b^x}$ using $b=0.01$ (for instance in the picture below) together with $y_0=x$ shows the three fixpoints. Looking at the curve for $y_1=b^x$ as well explains also, why there are three fixpoints.
[update - the proof for existence of three fixpoints]
The existence of two more crossing points where $t_k=ff(t_k)$ (fixpoints) is needed by two conditions.
Let's define $f(x)=b^x$, $ff(x)=b^{b^x}$ , and the identity function $g(x)=x$.
We know/observe, that $f(t)=t$ means $b=t^{1/t}$ and writing $u=\log(t)$ means $\log(b)= u / \exp(u)$ .
In the OP we have selected some $b < e^{-e}$ and from this we find a determination for $u$ and $t=\exp(u)$:
$$\begin{array} {}
\log(b)&=u / \exp(u) &\lt -1/ \exp(-1) = - 1 e^1 \\
\implies & -u \cdot \exp(-u) &> e \\
\implies & -u &> W(e)=1 \\
\implies & u &< -1 \\
\implies & | u| &> 1 \\
\end{array}$$
So the fixpoint for $f(x)$ is $t=\exp(u)< e^{-1} \approx 0.367879 $ and is of course a fixpoint for $ff(x)$ .
Let now, for example, $b=0.01$. Then $u =-1.28... $ and $t=0.278$.
We look at the first graph and want to argue that for $ff(x)$ besides of the fixpoint $t$ in the middle there need to be two more fixpoints (crossing points of the curve and the line), left and right to that fixpoint $t$.
(condition 1): $f(0)=b^0 = 1$ and thus $ff(0)=b$. The y-coordinate of the red curve at $x=0$ is $b$ and thus greater than the y-coordinate of $g(0)=0$ at $x=0$.
(condition 2): The function for the derivative of $ff(x)$ is $ff'(x) = \log(b)^2 b^x b^{b^x} $ At the fixpoint $t$ is thus the slope $ \log(b)^2 t^2 = (u/t)^2 t^2=u^2 >1 $ and thus must cross the line $g(x)$ from below. So the values for $ff(x)$ in the near left neighbourhood of $t$ must be smaller than the y-coordinate $g(x)$.
Because now at $x=0$ we have $ff(x)> g(x)=0$ but in the near left neighbourhood of $x=t$ we have $ff(t-\delta)< g(t-\delta)$ there must be some other point $t_1$ where $ff(t_{-1}) = g(t_{-1}) = t_{-1}$ and this is thus another fixpoint. Actually we find $t_{-1} = 0.013092... $.
The similar consideration for $x$ near $1$ gives a third fixpoint $t_1$ with $ff(t_1)=t_1 = 0.941488... $ Both fixpoints are moreover in the relation $f(t_{-1})=t_1$ and $f(t_1)=t_{-1}$
QED.
See this plot:
The red curve for the function $b^{b^x}$ crosses three times the blue $y=x$ line and by the grey curve for $b^x$ it should be easily recognizable why this must be thus.
Added: There exists this little
bonmot: "If you don't really understand, generalize". Things might even be clearer when more iterations are in the picture.
It suggests that even-height ($h=2k >0$) iterates should have three fixpoints and odd-height ($h=2k+1$) iterates should have one fixpoint.
