I have a permutation$f \in S_7$ with $|f| = 6$ and $h \in S_7$ with $|h| = 6$. I need to find a permutation, $j \in S_7$, such that $h=jfj^{-1}$.
$$ h = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 6 & 4 & 5 & 7 & 3 & 1 & 2 \end{pmatrix}\\ $$ $$ f = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 4 & 6 & 2 & 1 & 7 & 3 & 5 \end{pmatrix} $$ I tried writing both permutations as a product of transpositions and solving to find j but I'm not sure what to do.
A hint that might be useful to you is the following lemma.
Lemma: If $\sigma, \tau \in S_n$ with $\sigma = (a_1 \dots a_k)$ then $\tau\sigma\tau^{-1} = (\tau(a_1)\dots \tau(a_k))$.
Proving this isn't terribly difficult, and it can be leveraged with a little trick here. Note that in cycle notation your permutations may be written as $h = (16)(247)(35)$ and $f = (14)(263)(57)$. We want to find $j$ where \begin{align*} h &= jfj^{-1}\\ (16)(247)(35) & = j(14)(263)(57)j^{-1} \end{align*} The standard trick here is to multiply by $1 = j^{-1}j$ between each of the terms so as to get the conjugates of each cycle in the decomposition of $f$, namely $$ (16)(247)(35) = j(14)j^{-1}j(263)j^{-1}j(57)j^{-1} $$ then by applying the lemma we get $$ (16)(247)(35) = (j(1)j(4))(j(2)j(6)j(3))(j(5)j(7)). $$ With this you can match up the terms to get that $j = (1)(2)(375)(46) = (375)(46)$.
