What you are asking for is a conjugacy of the dynamical systems $g(x)=4x(1-x)$ and $h(x)=\sin(\pi x)$. Since $g$ and $h$ are both full unimodal maps of $[0,1]$, there will exist such a conjugacy, but it is very unlikely that you will be able to write it down explicitly. Also, it will certainly be continuous, but likely not absolutely continuous. 

You are asking that $f(g(x))=h(f(x))$. You are requiring that $f(0)=0$ and $f(1)=1$. Substituting $x=\frac 12$, you obtain $1=f(1)=h(f(\frac 12))$. Since $h^{-1}(1)=\{\frac 12\}$, it follows that $f(\frac 12)=\frac 12$. 
You can continue in a similar way to obtain the value of $f$ at further preimages of $\frac 12$. This defines $f$ on a dense set.

**EDIT**: As I mention above, it is very likely that $f$ is not absolutely continuous. David Speyer's argument below amplifies this. One key issue is that there are two important classes of invariant measure for maps like this: *absolutely continuous invariant measures* (ACIMs, that is, one that is absolutely continuous with respect to Lebesgue measure) and *measures of maximal entropy* (MMEs). From your post it seems as though you care about an ACIM for $h$.

For $g(x)$, the MME and ACIM coincide; this is something of a magical coincidence. A priori, it is highly unlikely that the ACIM and MME for $h(x)$ (assuming an ACIM exists) coincide. The conjugacy $f$ will always map the MME for $g$ to the MME for $h$; and so it seems very likely that this measure does not have a density. 

However: I did a computation that I expected would give a strong indication that the two measures are different. Namely, I numerically computed the Lyapunov exponent for $h$ at a point that I chose. Assuming that there is an ACIM, this would be equal to the measure-theoretic entropy of the ACIM. If this entropy is not $\log 2$ (the entropy of the MME), this would prove that the ACIM and MME do not coincide (in fact the converse holds also). To my surprise, the Lyapunov exponent for $h$ was (numerically) very close to $\log 2$, so who knows?!