Can I create a view on a vector?

What you can do, and probably want to do, is not use pointers at all - just sort the set of indices into locationsCompare, with the comparison function looking up the value in the original area. Easy-peasy with C++11:

template <typename T>
std::vector<size_t> get_sorted_positions(const std::vector<T> &v)
{
  std::vector<size_t> indices(v.size());

  std::iota(indices.begin(), indices.end(), 0); // indices now holds 0 ... v.size()-1
  std::sort(indices.begin(), indices.end(),
       [&v](size_t i_1, size_t i_2) { return v[i_1] < v[i_2]; }
  );

  return indices;
}

Notes:

• The only data getting mutated are the indices.
• Don't worry about returning a long vector; the compiler will use a move constructor, due to an optimization known as the NRVO.
• This code is mostly lifted from this answer, but the approach is basically folklore.
• You might also want to abstract away the fact that your input is a vector, and just take a reference to an arbitrary container (and return std::vector<typename Container::size_type>); or take a pair of iterators; or in C++20 - take any range.

The original vector must not be altered.

Consider to enforce this constraint by generating a vector of non owing pointers to const

template <class Container>
auto make_vector_of_const_pointers(Container& c)
{
    std::vector<typename Container::const_pointer> result;
    result.reserve(c.size());
    std::generate_n(std::back_inserter(result), c.size(),
                    [it = c.cbegin()]() mutable { return &(*(it++)); });
    return result;
}

See, e.g. here an example of usage, compared with a non const version.