Call two solutions equivalent if one is obtained from the other by permuting columns. I made some crude code, not well checked, and found these optimal solutions for vectors of length $n$.
$n=2$: best is 2, 2 nonequivalent solutions, example
10
01
$n=3$: best is 4, 2 nonequivalent solutions, example
110
101
011
111
$n=4$: best is 6, 5 nonequivalent solutions, example
1100
1010
1001
0110
1110
1101
$n=5$: best is 8, 265 nonequivalent solutions, example
11000
10110
10101
01011
01110
11100
11101
01111
$n=6$: best is 10, 80066 nonequivalent solutions, example
100000
010000
111000
100110
001110
010111
101101
101011
101111
110111
$n=7$: there are examples of size 13, not proved maximal yet, example
1000000
0100000
0010000
1101100
1100011
1011010
0001111
1010101
0110110
1110100
0111001
1111011
1101111
The method I'm using might be able to do $n=8$, but $n=10$ is impossible. If someone could verify the above are solutions, that would improve confidence in the results.