Who Is the Fairest of Them All?

Robert from Kings Ely sent in his solution:

The vector for the flags is $\frac{n-1}{n}$ of the vector of centres.

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:

Suppose the first scale factor is $k$ (so the second one is $\frac{1}{k}$). Let $\mathbf{x}$ denote the vector from the first centre of enlargement to the second. Then the required transformation is a translation by the vector $\frac{k-1}{k}\mathbf{x}$.

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 $\mathbf{a}$ 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 $k\mathbf{a}$. The vector from the second centre of enlargement to this image will be $k\mathbf{a}- \mathbf{x}$, so the vector from the second centre of enlargement to the final image will be $\frac{1}{k}\left(k\mathbf{a}-\mathbf{x}\right) =\mathbf{a}-\frac{1}{k}\mathbf{x}$. So the vector from the initial point to the final image will be $-\mathbf{a}+\mathbf{x}+\mathbf{a}- \frac{1}{k}\mathbf{x}=\frac{k-1}{k}\mathbf{x}$, as required.