Let us assume the word is $W = w_1w_2\ldots w_n$, of length $|W| = n$, and letters from some alphabet $w_k \in A$ for all $1\leq k \leq n$. From the given condition follows $|A|\geq 2$. Now, in order for a swap of letters to maybe create a periodic word, with (obvious $1 < p < n$) period length $p \mid n$, then each letter in the word must repeat (at least) $n/p \geq 2$ times. Among all letters of the word $W$ select one, say $a$, such that some difference $\delta$ between its consecutive apparitions is minimal in length. According to the given condition, we have $\delta > 1$, so $W = \ldots a w_1w_2\ldots w_{\delta} a \ldots$. Then swapping $aw_1$ (or $w_{\delta} a$) creates a word $W'$ with a strictly lesser difference $\delta'= \delta-1$, which is therefore unique. If $W'$ were periodic of $p\geq 3$ that will be a contradiction, so at most we will have $p=2$, with the two occurences of $a$ at closest distance lying in different periods. For this to occur $n$ must be even. But then, since $n\geq 6$, there will exist a subword $bcd$ in the second-half-word. Two of its possible swaps produce $cbd \neq bdc$ (since $b\neq c$), and so one of the resulting second-half-words will be different from the first-half word (which stays unchanged), thus the resulting word will be non-periodic. However, it is immediately seen that even for $n=4$ the result holds (not to mention smaller values of $n$), so the restriction $n>10$ was bogus.

I think your solution is not true we can have different periods length for different swaps so you can't swap cbd and bdc when you do this then P mabe changes and is not still 2 . But this works assum that two a's have the minimum length of two consecutive apparitions in all the word then swapping $a w_1$and$a w_2$ shows that p=2 and checking last letters of the word gives us contradiction.