Abstract Algebra: Equivalence Classes Under ~(conjugate) Relation Sorry. Don't know the latex.
Let $g=(1 \ 2 \ 3 \ 4 \ 5 \ 6 \ 7 \ 8)$ in $S_8$
Find a specific permutation $f$ in $S_8$ so that $fgf^{-1} =(1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4)$ (note: $fgf^{-1}$ is short for composition)
 A: Use the conjugation rule $fgf^{-1} = (f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8))$.
EDIT: Upon request, I give more details (which in principle are already given by Berci in the comments):
If f and g are any permutations, you can compute $fgf^{-1}$ by simply writing down $g$ as a product of disjoint cycles, and then applying $f$ entry-wise to this representation of $g$.
So in your case, $fgf^{-1} = (f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8))$.
Since your exercise states $fgf^{-1} = (1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4)$, we get $$(f(1)\; f(2)\; f(3)\; f(4)\; f(5)\; f(6)\; f(7)\; f(8)) = (1 \ 5 \ 6 \ 3 \ 8 \ 7 \ 2 \ 4).$$
Now we can simply read off that $f(1) = 1$, $f(2) = 5$, $f(3) = 6$, $f(4) = 3$, $f(5) = 8$, $f(6) = 7$, $f(7) = 2$, $f(8) = 4$.
That is (compare Berci's post) $$f = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\1 & 5 & 6 & 3 & 8 & 7 & 2 & 4\end{pmatrix}$$
or as a product of disjoint cycles
$$f = (2 \ 5\ 8\ 4\ 3\ 6\ 7).$$
A: Let $f$ map $1\mapsto 1,\ 2\mapsto 5,\ 3\mapsto 6,\ 4\mapsto 3,\ ...,\ 8\mapsto 4$.
