How to prove that the Permutation cipher is not CPA secure? Encryption scheme is called a permutation cipher if every plaintext string with letters
$a_1,\ldots, a_n$ is encrypted as $a_{\pi(1)}, \ldots, a_{\pi(n)}$ for some (not necessarily uniform) permutation $\pi$.

To break this cipher, one has to distinguish two messages based on their encryptions. I'm trying to solve this by playing a game. 

Two players A and U.

A: send a plaintext message to U (in this example A sends followashore)

U: Encrypt this message and send the cipher text to A


Input size of $\pi$: m = 6
$\pi$: $\begin{array}2 &amp;1 &amp;2 &amp;3 &amp;4 &amp;5 &amp;6 \\ &amp;3 &amp;5 &amp;1 &amp;6 &amp;4 &amp;2\end{array}$ 
Ciphertext: LEFOOLHEARSO


A: Changes this ciphertext by appending text $p$ of length $l$, which is a multiple of the length of the original plaintext. Let's say A sends LEFOOLHEARSOLEFOOLHEARSO (concatenated the ciphertext with itself)

U: Encrypt this message using the previous permutation?

I'm not sure where to go from here.
If $a_i = c$ for all $i$, then clearly $a_{\pi(i)} = c$ for all $i$ as well, regardless of the permutation $\pi$.

Thus, you can choose your two messages $a$ and $b$ such that, say, $a_i = 0$ and $b_i = 1$ for all $i$.  This will then also be true for their encryptions under any permutation cipher, allowing you to trivially distinguish the encrypted messages with 100% certainty.