For $\O\colon=(0,1)$ and $\theta(x)=4x(1-x)$ let $\mu$ be the probability measure with density $f(x)=\pi^{-1}(x(1-x))^{-1/2}$. Then $\mu$, $\d_0$ and $\d_{3/4}$ are $\theta$-invariant. Moreover, put $F(x)=(\sin(\pi x)/2)^2$, then
$$
F^{-1}\circ\theta\circ F(x)
=\left\{\begin{array}{cl}
2x&\mbox{if $x < 1/2$}\\
2(1-x)&\mbox{if $x\geq1/2$}
\end{array}\right.
$$
For all $t\in(0,1)$ we have $[\theta\leq t]=(0,\vp(t)]\cup[1-\vp(t),1)$ where $\vp(t)=(1-\sqrt{1-t})/2$. The distribution function of $\theta$ is given by
$$
D(t)\colon=\mu(\theta\leq t)=2\int_0^{\vp(t)}f(x)\,dx,
$$
i.e. $D^\prime(t)=2\vp^\prime(t)f(\vp(t))$, which is readily seen to be equal to $f(t)$ and thus $\theta$ preserves the measure $\mu$. Finally $0$ and $3/4$ are fixed point of $\theta$; hence both $\d_0$ and $\d_{3/4}$ are $\theta$-invariant. The last statement is left as an exercise.