Let $(c_{1} \dots c_{r})$ and $\gamma$ be permutations, prove that $$(c_{1} \dots c_{r})^\gamma = (c_{1}\gamma \dots c_{r}\gamma ).$$
What I've tried is:
Let $c$ be a digit, if $c=c_j\gamma$ for some $j$, we have $c(c_{1}\dots c_{r})^\gamma=c\gamma^{-1}(c_{1}\dots c_{r})\gamma=c_j(c_1\dots c_r)\gamma=c_{j+1}\gamma$ and I don't know if I'm going in the right way.
Thanks in advance.
$c=c_j\gamma$ is not necessarily true unless $c\in \{c_1,\dots,c_r\}$.
Let $\sigma=(c_{1} ... c_{r})^\gamma$. We want to show that
(1) $(c_i\gamma) \sigma=(c_{i+1}\gamma)\sigma$ for $i=1,\dots,r-1$.
(2) $(c_r\gamma)\sigma=(c_1\gamma)\sigma$.
(3) $k\sigma=k$ for $k\notin \{c_1\gamma,\dots,c_r\gamma\}$.
From what you had done, I believe that you can complete (1) and (2).
To prove (3), note that since $k\notin \{c_1\gamma,\dots,c_r\gamma\}$, we have $k\gamma^{-1}\notin \{c_1,\dots,c_r\}$. Thus,
$$k\sigma=(k\gamma^{-1})(c_1\dots c_r)\gamma=k\gamma^{-1}\gamma=k.$$
This completes the proof.
