Sorting a vector in descending order within two ranges

Your comparison function is wrong since the values you get as first and second are the elements of the std::vector. Therefore, there is no need to use them as indices. So, you need to change

return indices[first] > indices[second];

to

return first > second;

Now, regarding the problem you try to solve...

You can leave 3, 4, 5 and 6 out of comparison with other elements and still compare it with each other:

std::sort(
    indices.begin(), indices.end(),
    [](int first, int second) -> bool {
        bool first_special = first >= 3 && first <= 6;
        bool second_special = second >= 3 && second <= 6;
        if (first_special != second_special)
            return second_special;
        else
            return first > second;
    }
);

Demo

Functions from the standard algorithms library like iota, sort, find, rotate and copy would make your life easier. Your example comes down to:

#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
#include <iterator>


int main()
{
  std::vector<int> indices(15);
  std::iota(indices.begin(), indices.end(), 0);
  std::sort(indices.begin(), indices.end(), std::greater<>());

  auto a = std::find(indices.begin(), indices.end(), 6);
  auto b = std::find(indices.begin(), indices.end(), 3);
  std::rotate(a, b + 1, indices.end());

  std::copy(indices.begin(), indices.end(), std::ostream_iterator<int>(std::cout, "\n"));
  return 0;
}

Output:

14
13
12
11
10
9
8
7
2
1
0
6
5
4
3

---

@TedLyngmo in the comments makes the good point that it could/should be improved with:

auto a = std::lower_bound(indices.begin(), indices.end(), 6, std::greater<int>{});
auto b = a + 4;