The behaviour of a solid object under heating is very easy to state:
Every linear dimension of the object expands the same way:$$\Delta l=k\times l\times \Delta T$$
Where $\Delta l$ is the change in some linear dimension, $l$ is the original value of the dimension, $k$ is the coefficient of linear thermal expansion for the material in question, and $\Delta T$ is the change in temperature. The choice of Fahrenheit or Celsius scales must be the same for both $k$ and $\Delta T$
This applies to any dimension of an object: length, width, thickness, distance between two scratches on the surface, diameter of any hole, width of any ring etc.
For example, one way of fixing a disc on a shaft (say a train wheel on an axle) is to drill a hole in the disc, slightly smaller than the shaft. Heat the disc until the hole expands just enough, slide over the shaft to the correct position (quickly!), and allow to cool. The disc will be locked onto the shaft...
Specifically, in regards to your question, the entire aluminium ring will be stretched out in every direction, and its proportions will not change—the final shape will look exactly like the original shape, just a little bit bigger.