The answer is yes, here is a construction. 0000000000000000 ____________________ 1111111111111111 ____________________ 0101010101010101 1010101010101010 ____________________ 0111011101110111 1110111011101110 1101110111011101 1011101110111011 ____________________ 1000100010001000 0001000100010001 0010001000100010 0100010001000100 ____________________ 0011001100110011 0110011001100110 1100110011001100 1001100110011001 ____________________ 0110100101101001 1101001011010010 1010010110100101 0100101101001011 1001011010010110 0010110100101101 0101101001011010 1011010010110100 ____________________ 0011110000111100 0111100001111000 1111000011110000 1110000111100001 1100001111000011 1000011110000111 0000111100001111 0001111000011110 ____________________

Let $S$ be the set of binary strings of length $5$, i.e. $S = \{0, 1\}^5.$ We will say that $s_1, s_2 \in S$ satisfy the relation $s_1 \rightarrow s_2$ if the last four digits of $s_1$ are the first four digits of $s_2.$ Now, a row of the table corresponds to $s_1, s_2, \cdots, s_{16} \in S$, satisfying $s_1 \rightarrow s_2 \rightarrow s_3 \rightarrow \cdots \rightarrow s_{16} \rightarrow s_1$. From here, it would suffice to partition the elements of $S$ into cycles of lengths divisible by $16$. In this way, if we had a cycle $s_1 \rightarrow s_2 \rightarrow \cdots \rightarrow s_t \rightarrow s_1$ for $t | 16$, we could let our first row be such that it reads $s_1, s_2, \cdots, s_t$ each $\frac{16}{t}$ times, and cyclically shift this first row by one $t-1$ times to get $t-1$ other row. (These groups of cyclically shifted rows are delineated in the answer with _____)

0000010001100111 1111101110011000 + their cyclic shifts

The following numbers work: 000000|00000|00000 000011|00001|00001 000101|00010|00010 000110|00011|00011 001001|00100|00100 001010|00101|00101 001100|00110|00110 001111|00111|00111 010001|01000|01000 010010|01001|01001 010100|01010|01010 010111|01011|01011 011000|01100|01100 011011|01101|01101 011101|01110|01110 011110|01111|01111 100001|10000|10000 100010|10001|10001 100100|10010|10010 100111|10011|10011 101000|10100|10100 101011|10101|10101 101101|10110|10110 101110|10111|10111 110000|11000|11000 110011|11001|11001 110101|11010|11010 110110|11011|11011 111001|11100|11100 111010|11101|11101 111100|11110|11110 111111|11111|11111 This clearly works upon inspection.

Call a $2^n \times a$ table constructed on a cylinder "good" if we can fill the cells with 0 or 1 so that any $2^n \times n$ table constructed by $n$ consecutive columns is varied, where $a \ge n$, and we extend the definition of varied tables to $2^n \times n$ tables rather than just $32 \times 5$ tables. It is easy to find that if we can construct a good $2^n \times a$ table and a good $2^n \times b$ table, then by joining them together (as shown in the above solution with "|"), we can construct a good $2^n \times (a+b)$ table. Clearly, we can construct a good $2^n \times n$ table. We can show that we can also construct a good $2^n \times (n+1)$ table, by induction on n. Then, by putting together 2 $32 \times 5$ tables and 1 $32 \times 6$ table, we have a good $32 \times 16$ table, as desired.

0000000000000000 1111111111111111 1000100010001000 and cyclic shifts (4) 0111011101110111 and cyclic shifts (4) 1100110011001100 and cyclic shifts (4) 0101010101010101 and cyclic shifts (2) 1001011010010110 and cyclic shifts (8) 1100001111000011 and cyclic shifts (8)