In a lambda, does a by-value capture of a reference copy the underlying object?

If a variable of reference type is captured in a lambda by value, is the referenced object copied or is it captured by reference?

Small sample with question:

#include <iostream>

struct Test {
  int a;
};

void testFunc(const Test &test) {
  auto a = [=] {
    // is 'test' passed to closure object as a copy
    // or as a reference?
    return test.a;
  } ();
  std::cout << a;
}

int main() {
  Test test{1};
  testFunc(test);
}

By value. Compilable example:

class C
{
public:
	C()
	{
		i = 0;
	}

	C(const C & source)
	{
		std::cout << "Copy ctor called\n";
		i = source.i;
	}

	int i;
};

void test(C & c)
{
	c.i = 20;

	auto lambda = [=]() mutable {

		c.i = 55;
	};
    lambda();

	std::cout << c.i << "\n";
}

int main(int argc, char * argv[])
{
	C c;
	test(c);

	getchar();
}

Result:

Copy ctor called
20

I guess, that this paragraph of the C++ standard applies:

5.1.2 Lambda expressions

(...)

14. An entity is captured by copy if it is implicitly captured and the capture-default is = or if it is explicitly captured with a capture that does not include an &. For each entity captured by copy, an unnamed nonstatic data member is declared in the closure type. The declaration order of these members is unspecified. The type of such a data member is the type of the corresponding captured entity if the entity is not a reference to an object, or the referenced type otherwise. [ Note: If the captured entity is a reference to a function, the corresponding data member is also a reference to a function. —end note]

That actually makes sense - if local variables are passed by value and parameter passed by reference "acts" as a local variable in function, why would it be passed by reference instead of value?