Concatenating multiple FFTs
It is not possible to simply concatenate FFTs to obtain a larger one (that would be a neat trick if it worked). It is possible to combine several smaller point FFT's to create a larger one. To see this, let's start with the original equation,
$$ X[k] = \sum_{n=0}^{N-1} x[n] \exp\left(-\frac{j 2 \pi k n}{N} \right). $$
Now, consider that we have used this formula on two sequences already to obtain two FFTs of length $\frac{N}{2}$,
$$ X_0[k] = \sum_{n=0}^{N/2-1} x[2 n] \exp\left(-\frac{j 2 \pi k n}{N/2} \right) $$
and
$$ X_1[k] = \sum_{n=0}^{N/2-1} x[2 n + 1] \exp\left(-\frac{j 2 \pi k n}{{N/2}} \right). $$
Now, we have two FFTs of length $\frac{N}{2}$, and we want to take the information contained in them to form a larger FFT. Specifically, we would like to form
$$ Y[k] = \sum_{n=0}^{N - 1} x[n] \exp\left(-\frac{j 2 \pi k n}{N} \right). $$
To see how to obtain this, we decompose the summation into two parts, one containing even terms and one containing odd terms, 
$$ Y[k] = \sum_{n=0}^{N/2 - 1} x[2 n] \exp\left(-\frac{j 2 \pi k \left( 2 n\right)}{N} \right) + \sum_{n=0}^{N/2 - 1} x[2 n + 1] \exp\left(-\frac{j 2 \pi k \left(2 n + 1\right)}{N} \right). $$
Some simple transformations turn this into,
$$ Y[k] = \sum_{n=0}^{N/2 - 1} x[2 n] \exp\left(-\frac{j 2 \pi k n}{N/2} \right) + \exp\left(-\frac{j 2 \pi k}{N}\right) \sum_{n=0}^{N/2 - 1} x[2 n + 1] \exp\left(-\frac{j 2 \pi k n}{N/2} \right). $$
And this yields our answer. To obtain an FFT that is twice the length of two smaller FFTs, we simply apply a twiddle factor to the later samples and combine the two smaller FFTs via addition. Explicitly, based on the notations above, we can write
$$ Y[k] = X_0[k] + \exp\left(-\frac{j 2 \pi k}{N}\right) X_1[k]$$
where the index $k$ is taken modulo-$\frac{N}{2}$ or else $X_0[k]$ and $X_1[k]$ are considered periodic functions with period $\frac{N}{2}$.

You can arrive at a similar formula through a decimation in frequency approach. I leave the details for that to you or someone else.

Edit: If you are dead set on using $N=5120$ as your output FFT size, then you can use the following variation. First compute 5 FFTs of length 1024. You can compute these by taking every fifth number from the original sequence and applying the FFT from the library. For notational convenience, let's define these as
$$ X'_p[k] = \sum_{n=0}^{N/5-1} x[5n+p] \exp\left(-\frac{j 2 \pi k n}{N/5}\right) \:\: \mbox{for $p = 0, 1, 2, 3, 4$}.$$
Now, combine these components with the equation
$$ X[k] = \sum_{p=0}^{4} \exp\left( -\frac{j 2 \pi p k}{N} \right) X'_p[k].$$
Again, the indices should be considered modulo-$\frac{N}{5}$ in this case.