How to conditionally define a lambda?

Why this fails is because the lambdas are of a different type. That's natural, their operator() have different definitions. Which means you want your following code to work with two different types. And the C++ way of making code work with different types is using templates.

Convert the code using getJ to a function template (it can be local to your implementation file), like this:

template <class G>
void salt_impl_(Mat mat, unsigned long long n, default_random_engine &generator, G getJ)
{
    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())
    {
        cols *= rows;
        rows = 1;
    }

    uchar * const data = mat.data;

    uniform_int_distribution<uint> randomCol(0, cols - 1);

    for (unsigned long long counter = 0; counter < n; counter++)
    {
        uint i = randomCol(generator);
        uint j = getJ();

        uint index = channels * (cols * j + i);
        for (uchar k = 0; k < channels; k++)
            data[index + k] = 255;
    }
}


void salt_(Mat mat, unsigned long long n)
{
    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())
    {
        cols *= rows;
        rows = 1;
    }

    default_random_engine generator;
    uniform_int_distribution<uint> randomRow(0, rows - 1);

    if (rows == 1)
      salt_impl_(mat, n, generator, []() {return 0; });
    else
      salt_impl_(mat, n, generator, [&]() {return randomRow(generator); });
}

Feel free to reduce the initial-part duplication between the function and the template by passing more parameters, making them members of a class, or something similar.

Also note that the non-trivial lambda must capture the variables which it accesses (randomRow and generator). I did this using the universal by-reference capture [&] in the code above.

my code with this lambda does not compile with the following red squiggled text

You cannot use randomRow inside the body of the lambda expression without capturing it beforehand, as the generated closure object needs to have access to it.

Even if you were to use [&randomRow], the code would still fail to compile as every lambda expression produces a closure of unique type, even if the lambda expressions are exactly the same.

You can turn the problem on its head to avoid any overhead and achieve what you want - create a function that takes the lambda you want to invoke:

template <typename F>
void saltImpl(F&& getJ, /* ... */)
{
    uchar * const data = mat.data;

    for (unsigned long long counter = 0; counter < n; counter++)
    {
        uint i = randomCol(generator);
        uint j = rows == 1 ? 0 : randomRow(generator);
        //uint j = getJ();

        uint index = channels * (cols * j + i);
        for (uchar k = 0; k < channels; k++)
            data[index + k] = 255;
    }
}

Usage example:

void salt_(Mat mat, unsigned long long n)
{
    const uchar channels = mat.channels();
    uint cols = mat.cols;
    uint rows = mat.rows;

    if (mat.isContinuous())
    {
        cols *= rows;
        rows = 1;
    }

    default_random_engine generator;
    uniform_int_distribution<uint> randomRow(0, rows - 1);
    uniform_int_distribution<uint> randomCol(0, cols - 1);

    if (rows == 1)
    {
        saltImpl([]{ return 0; }, /* ... */);
    }
    else
    {
        saltImpl([&]{ return randomRow(generator); }, /* ... */)
    }
}