Question: If I have a list of 6 options to choose from and a selector that has 8 possible values, can that selector be reduced to evenly select from the 6 options?

I'm using 3 bits, so a value between 0 and 8 exclusive and I need to select from 6 options. My understanding is that I have more information than I need, but I don't know how to construct a map to the lower information state.

My only idea so far is to use the selector value to seed a pseudo-random number, but is there enough information in the seed to assume an even distribution?

Answer: 
It follows from the assumption that for every $y$ in the image of $f$ we have$$f(y)=\frac{1}{y}.$$Now, as you already found out yourself, both $9$ and $1/9$ are in the image. As $f$ is assumed to be continuous, it follows that $5$ is also in the image, and so,$$f(5)=\frac{1}{5}.$$
You have calculated $f(9)=1/9$. So consider the interval $[9,10]$. Since $f$ is continuous, by intermediate value theorem, you can find $z$, which lies in $[9,10]$ such that $f(z)=5$.

Now $f(z)\cdot f(f(z))=5 \cdot f(5)=1$. Then $f(5)=1/5.$