How would one go about proving the claim in the title?
I see that if $\alpha,\beta$ are $n$-cycles and $\alpha,\beta$ permute $A,B\subset \{1,\dots, n+2\}$ respectively then $\overline{A\bigcap B}$ has size $0,\, 2$ or $4$. If $0$ then there is an odd permutation $\alpha=(a_1\cdots a_n)\to \beta=(b_1\cdots b_n)$ defined by $a_i\mapsto b_i$. This permutation is because $\{a_i:\;a_i=b_i\}$ must have even size.
I cannot see why it is still true if $\overline{A\bigcap B}$ has size $2$ though. Am I even going down the right route?
In the case you give, both cycles permute the same $n$ elements, and are conjugate within the group $S_n$ on those elements. If they are conjugate in $S_n$ by an even permutation you are done. If the permutation is odd, then there are two more elements in $A_{n+2}$ to create a transposition, which converts your odd permutation to an even one, and because $\tau^2=1$ for a transposition it doesn't affect the conjugation.