I propose you a simple strategy to solve that, based on looking for rectangles intersections.
The key for that is to create a simplified representation of your images with a set of rectangles laid out properly as bounding boxes of the different part of you image (like you would build your image with legos). For better performance use a small set of rectangles (a few big legos), for better precision use a biggest number of rectangles to precisely follow the image outline.
Your problem becomes equivalent to finding an intersection between rectangles. Or to be more precise to find wether at least one vertex of the rectangles of object A is inside at least one rectangle of object B (CGRectContainsPoint) or if rect intersects (CGRectIntersectsRect).
If you prefer the points lookup, you should define your rectangles by their 4 vertices then it is easy when you rotate your image to apply the same affine transform (use CGPointApplyAffineTransform) to your rectangle vertices to have the coordinates of your points after rotation. But of course you can lookup for frame intersections and represent you rectangle using the standard CGRect structure.
You could also use a CGPath (as explained in another answer below) instead of a set of rectangles and look for any vertex inside other path using CGPathContainsPoint. That would give the same result actually but probably the rectangles approach is faster in many cases.
The only trick is to take one of the objects as a reference axis. Imagine you are on object A and you only see B moving around you. Then if you have to rotate A you need to make an axis transform to always have B transform relatively to A and not to the screen or any other reference. If your transforms are only rotation around the object centre then rotating A by n radians is equivalent to rotating B by -n radians.
Then just loop through your vertices defining object A and find if one is inside a rectangle of object A.
You will probably need to investigate a bit to achieve that but at least you have some clues on how to solve that.