Isomorphism composed with a certain conjugation fixes (some) 2-cycles in $S_n$

Question

If $\psi: S_n \to S_n$ is an isomorphism and $\psi(12)=(ij), \psi(13)=(ik)$ for $k \not={i,j}$, show that there exists $g \in S_n$ such that $\psi (g(12)g^{-1})=(12)$ and $\psi(g(13)g^{-1})=(13)$.

I have the following results at my disposal:

1. Let $cl(c)$ denote the conjugacy class of $c$. Then $\psi(cl(c))=cl(\psi(c))$ and $|cl(c))|=|cl(\psi(c))|$.

2. For $n \not= 6$, $|cl(\sigma)|=|cl((12))|$ iff $\sigma$ is a 2-cycle.

EDIT: I have some basic ideas but I am not sure if it is correct. Take $g \in S_n$ such that $g(ij)g^{-1}=(12)$. We know such $g$ exists because $(ij)$ and $(12)$ are conjugates in $S_n$. Consider $\psi^{-1}(g) \in S_n$. Then $\psi(\psi^{-1}(g)(12)\psi^{-1}(g)^{-1}) = g(ij)g^{-1}=(12)$. So we have found such $g$ for $(12)$. However, now I am having trouble showing that the same $g$ works for $(13)$....

Answer 1

You can pick $h$ so that $h(i)=1$, $h(j)=2$, $h(k)=3$ and let $g=\psi^{-1}(h)$.