Robert from Kings Ely sent in his solution:

Unfortunately no-one has sent in an explanation of how to arrive at this answer, so see if you can follow it through using this diagram:

To see this, consider a single point on the flag. (If we show that the required transformation for a single point is the given translation, then the same will apply to the flag.) Let $a$ denote the vector from the first centre of enlargement to the point. Then the vector from the first centre of enlargement to the image of the point under the first enlargement will be $ka$. The vector from the second centre of enlargement to this image will be $ka-x$, so the vector from the second centre of enlargement to the final image will be $\frac{1}{k}(ka-x)=a-\frac{1}{k}x$. So the vector from the initial point to the final image will be $-a+x+a-\frac{1}{k}x=\frac{k-1}{k}x$, as required.