15. (1) Starting from the digit 0, repeatedly applying the transformation $T: 0 \rightarrow 01, 1 \rightarrow 10$, we get $T(0)=01, T^{2}(0)=T(01)=$ $0110, \cdots$, thus we obtain the Morse-Thue sequence in another way, where $T^{n}(0)$ represents the first $2^{n}$ digits of the sequence. Applying the transformation $T$ to the entire sequence leaves it unchanged, which makes (1) almost obviously true, and (2) is the same.
(3) We notice that the adjacent terms $x(2 n)=x(n)$ and $x(2 n+1)=1-x(n)$ are always different. If the sequence eventually becomes periodic, and let $x(n)$ be an element within a period of the sequence, then $n+1$ must not be a multiple of the period (otherwise $x(2 n+1)=x(n)$, a contradiction). Similarly, it can be proven that $n+2, n+3, \cdots$ are also not multiples of the period, but this is impossible.
(4) This conclusion is correct because the $(n+1)$-bit 0-1 sequence is formed by adding a 1 and possibly some 0s in front of the binary numbers up to $n$ bits.