You seem to think that those two lambdas have the same type, but that is not true. Each one creates its own type:

#include <functional>
#include <type_traits>
#include <iostream>

int main() {
  auto test = []{};
  auto test2 = []{};
  std::cout << std::is_same< decltype( test ), decltype( test2 ) >::value << std::endl;
  return 0;
}

will print 0. Of course the error message you are getting from the compiler could be a little bit clearer in this regards…