Sort an array with small numbers while having secondary array with initial indexes

First, it seems much better as the comments suggest to just use a structure, containing both elements and their indices: struct item { int value; int index; }. You then sort them using any sort you like using item.value only, and you have indices then sorted for you.

However if you specifically want to use counting sort on given data structures, I'd probably modify this counting sort algorithm from simply counting to storing indices instead. For example, where you normally would store a std::vector counters or std::map<int, int> to keep the number of each value in the array, you can instead use a vector/map of vectors/sets. Here an element at index 0 would store all indices of the original array, where the value was 0. Example using just vectors:

say original_array contains {0,3,5,1,0,2,4}

your new structure:

std::vector<std::vector<int>> counters(7); // one vector for each possible value from 0 to 6

counting:

for (int index = 0; index < original_array.size(); ++index)
{
    const int& value_at_index = original_array[index];
    counters[value_at_index].push_back(index);
}

in the end your counters is going to contain all the information you need. For each element of the counters array, the number of sub-elements is its frequency, and subelements themselves give you indices of the original items. Also note a property that this is a stable sort, i.e. elements of equal value preserve their indices order. An example of what's inside indices after this:

0:0,4
1:3
2:5
3:1
4:6
5:2
6:-

Hope this helps!