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?!