So it is not hard to see that the conjugacy class of $(123)(456)$ in $S_7$ has cardinality $\frac{7!}{3\cdot 3\cdot2}=280$.
But for instance to go from $(123)(456)$ to $(132)(456)$ you have to conjugate by $(23)$ which is not in $A_7$ so $(132)(456)$ is not in our class (call it $C$). Same for $(123)(465)$. But $(132)(465)\in C$ because $(23)(56)\in A_7$.
I could probably find all the elements of $C$ by trying out different permutations but I feel like there is a better way...
We know that $|A_7|={|S_7|\over2}$ and conjugating $(123)(456)$ by any permutation in $S_7\supset A_7$ gives a permutation of the type $(abc)(def)~$ with $a,...,f\in\{1,...,7\}$ and $a,b,...,f$ are disjoint because otherwise the "permutation" by which we conjugate wouldn't be bijective.
So I'm tempted to say that $|C|=|\bigcup\limits_{~a\in A_7}\{(~a(1)a(2)a(3)~)~(~a(4)a(5)a(6)~)~\}|=\frac{280}{2}$ but that's probably false.
A hint would be appreciated
Hint: If you conjugate $(123)(456)$ by $(14)(25)(36)$, an odd permutation, you get $(123)(456)$ back. Therefore you also get $(132)(456)$ from $(123)(456)$ by conjugating it by an even permutation.