This makes no difference because the iterator returns pairs of references:

Then my understanding of auto must be rusty, I thought auto x = f() never resolves to a reference?

It doesn't. Here, f() returns a pair of references. auto doesn't deduce to a ref for the pair itself. But it preserves the refs of the pairs contents. Have a look at this:

#include <iostream>
#include <functional>
#include <utility>

std::pair<int, int&> f(int &x) {
    return std::pair{0, std::ref(x)};
}

int main() {
    int x = 2;
    auto [_, y] = f(x);
    y = -1;
    std::cout << x << '\n';
}

This prints -1.

👍