conjugacy in symmetric group I was reading Dummit and Foote and encountered the following statement: any two elements in $S_n$ are conjugate if and only if they have the same cycle types.
However, I am able to produce a counter example:
Let $(1 2 3)$ and $(4 5 6) (7 8)$ be in $S_{10}$, then $(4 5 6) (7 8) =(1 3 2) (4 5 6) (7 8) (1 2 3)$, which show that these two are conjugate.
What am I misunderstanding here?

Of course a permutation is conjugate to itself.  It has the same cycle type as itself, as well.  You could just as well have conjugated by the identity.

Conjugation takes $k$-cycles to $k$-cycles:  $\pi^{-1}(a_1\dots a_k)\pi=(\pi(a_1)\dots\pi(a_k))$.
Also, conjugation is a homomorphism.   So, under conjugation, a product of cycles is the product of the conjugates.  To finish, use that any permutation has a representation as a product of disjoint cycles.
Thus there can be no counterexample.
