Prove that $\tau \sigma \tau ^{-1}$ depends only on the lateral class of $\tau$ with respect to $Z(\sigma)$

Question

I want to prove that, given $\sigma, \tau, \tau ' \in S_n$, we have that:

$\tau ' \in \tau H \Longleftrightarrow \tau \sigma \tau ^{-1}= \tau ' \sigma (\tau ')^{-1}$

Where $H=Z(\sigma)$ is the centralizer of $\sigma$. I proved $(\Longrightarrow)$:

If $\tau ' \in \tau H$, then $\tau '=\tau \rho$ where $\rho \in H$. This means that $\tau ' \sigma (\tau ')^{-1}=\tau \rho \sigma \rho ^{-1} \tau ^{-1}=\tau \sigma \tau ^{-1}$.

I don't know what to do for ($\Longleftarrow$).

Answer 1

If $\tau\sigma\tau^{-1} = \tau'\sigma(\tau')^{-1}$, then premultiplying both sides by $(\tau')^{-1}$ and postmultiplying by $\tau$ gives $$(\tau')^{-1}\tau\sigma = \sigma(\tau')^{-1}\tau.$$ Thus $(\tau')^{-1}\tau\in H = Z(\sigma)$.