Short answer: put n=426 in the calculator, but beware of assumptions detailed in long answer.
Long answer:
I assume here that your null hypothesis is that correlation between a and b is the same than correlation between c and d.
The first thing to do would be to test for correlation between a-b correlation and c-d correlation. If they are uncorrelated maths became a lot simpler, although if you can estimate correlation between both correlation r-z transform could still be useful. From here I will assume that a-b correlation is not correlated to c-d correlation.
Assuming that a and b are close enough to a normal distribution, every 50 sample a-b correlation, once transformed, will result in a value distributed normally with mean ${1 \over 2}\ln\left({{1+\rho} \over {1-\rho}}\right)$ and variance $\frac{1}{N-3}$ (with $N=50$). Under the null hypothesis that a-b and c-d have the same correlation, transformation of c-d correlations will also be normally distributed with the same mean and variance.
The average of correlations of 9 samples will be normally distributed with the same mean and $1/9$ of the variance. That is:
$$var(mean(correlation(a,b)))=var(mean(correlation(c,d)))=\frac{1}{9}·\frac{1}{50-3}=\frac{1}{423}$$
From here you could end your test by hand, since the difference of both means will be normally distributed with mean 0 and variance double of each one:
$$var(mean(correlation(a,b))-mean(correlation(c,d)))=2·\frac{1}{423}$$
And you can use that to compute the p-value using difference of z values.
Anyway, we could also use the calculator, but we'll need to tell it a little lie.
We need to notice that the calculator just uses N to compute the variance of z values, because z values are just function of r, not N. Therefore, if the calculator is given the means of the correlations it will compute z values correctly.
After getting the two z values the calculator just finds the p-value assuming that both z values are normally distributed with the same mean and variance $\frac{1}{N-3}$. That's just what we want it to do, except that we know that variance is not $\frac{1}{N-3}$ but $\frac{1}{423}$ (because we are not giving the calculator two correlation but two means of 9 correlations). Here comes our little lie: we tell the calculator that $N=426$ and it will use variance $\frac{1}{426-3}=\frac{1}{423}$ and return the correct p-value.
In this answer I made several assumptions that I'd like to discuss:
- Independence of $correlation(a,b)$ and $correlation(c,d)$: if they are correlated, both z values will be correlated too, and it will need to be taken in account in the distribution of difference of p-values. The challenge might be estimate correlation between correlations (because of small sample of correlations: 9) and to transform that correlation to correlation between z values (that might not be straightforward). Anyway, unaccounted lack of independence would produce a lower variance and and higher p-value. Therefore, if you reject null hypothesis assuming independence, you would reject it even if lack of independence is taken in account; the opposite is not true.
- Null hypothesis is that all correlations are the same: For me, that is the most obvious interpretation of the question. However, another possible interpretation is that the mean of $correlation(a,b)$ equals the mean of $correlation(c,d)$. That would just be a two means test that could be solved using a t-test if we can made some assumptions on the distribution on correlations or otherwise can be attempted using a non parametric test.
