Two codes $C_1, C_2 \subseteq A^n$ are called equivalent (notation: $\sim$) if there are permutations $\pi \in Sym(A)$ and $\sigma_1, \dots, \sigma_n \in S_n$ such that
$$C_2 = \{(\sigma_1(a_{\pi(1)}), \dots, \sigma_n(a_{\pi(n)})) \mid (a_1, \dots, a_n ) \in C_1\}$$
Intuitively, this means that we can permute the order of the symbols in our code words, and subsequently permute the alphabet $A$ on every single position on everyy word.
The text I'm reading mentions without proof that $\sim$ defines an equivalence relation, but I do not succeed in proving this.
In particular, I'm stuck at showing that
$$C_1 \sim C_2 \implies C_2 \sim C_1$$
I tried with a little example:
if we have the code $C_1 := \{00000,01101,10110,11011\} \subseteq \{0,1\}^5$ and we perform $\pi = (12)(345)$ and $\sigma_3 = \sigma_4 = 1$, $\sigma_1 = \sigma_2 = \sigma_5 \neq 1$, then we get the code $C_2 = \{11001,01111,10010,00100\}$
To get back to $C_1$, we have to apply $\pi' = (21543)$ and $\sigma_2' = \sigma_3' = 1$ and $\sigma_1' = \sigma_4' = \sigma_5' \neq 1$. I don't see a relation between the permutations $\pi, \sigma_i$ and $\pi', \sigma_j$ that can help me.
How can I show that this relation is symmetric?
Answering my own question:
It is easy to check that (show that the right side is contained in the left side and then use that both sides have equal finite order)
$$C_1 = \{(\sigma^{-1}_{\pi^{-1}(1)}(a_{\pi^{-1}(1))}), \dots, (\sigma^{-1}_{\pi^{-1}}(a_{\pi^{-1}(n))})\mid (a_1, \dots, a_n) \in C_2\}$$
and the symmetry follows.