When does multiple OTP encryption become insecure if new keys are permuted? ​
This is an inconsequential variation on a two-time pad and falls just as readily.

Say your pad is $n$ symbols long.  Fix a permutation $\pi \in S_n$ of $\{1,2,\dots,n\}$—this is determined by your $\operatorname{permute}$ operator.  If the pad is $(p_1, p_2, \dots, p_n)$ and the $i^{\mathit{th}}$ message is $m_i = (m_{i,1}, m_{i,2}, \dots, m_{i,n})$, then the $i^{\mathit{th}}$ ciphertext $c_i = (c_{i,1}, c_{i,2}, \dots, c_{i,n})$ where $$c_{i,j} = m_{i,j} + p_{\pi^{i-1}(j)}.$$  Here $\pi^i(x) = \pi(\pi(\cdots \pi(x)\cdots))$ is the $i$-fold iteration of $\pi$, with $\pi^0(x) = x$.

So far this is just a restatement of your system, with specific names given to each symbol in the message, pad, and ciphertext.

If I learn the ciphertext $c_1$ of a single known message $m_1$, then I can trivially recover your pad by $$p_j = p_{\pi^0(j)} = c_{1,j} - m_{1,j},$$ and then decrypt the next ciphertext $c_2$ or any other by simply permuting the pad $(p_1, p_2, \dots, p_n)$ with $\pi$.

Obviously, if I don't know $m_1$ exactly but can narrow it down to a small set of options, then I can narrow down the $p_j$ to a small set of options too, and thus narrow down $m_2$, and so on.

In general, if the pad for the first message is $p$, the pad for the second message is $F(p)$, the pad for the third is $F^2(p)$, etc., where $F$ is a fixed function of pads (in this case, $F(p_1, \dots, p_n) = (p_{\pi(1)}, \dots, p_{\pi(n)})$), the stream cipher you get is completely broken.  When $F$ is a permutation with an easily computed inverse, as it is here, a known plaintext anywhere among the messages is enough to decrypt all other messages, even prior ones.

For the same reason, $\operatorname{MD5}(k) \mathbin\| \operatorname{MD5}^2(k) \mathbin\| \cdots$ makes for a terrible stream cipher even though $\operatorname{MD5}(k \mathbin\| 1) \mathbin\| \operatorname{MD5}(k \mathbin\| 2) \mathbin\| \cdots$ is basically fine.