ShapeLayer rectangle size and trim path start, end, offset, expression to adjust with rectangle siz

Yeah, you probably need to correlate the absolute path length to the values generated by the End and Start percentages somehow to get some real pinning. Should not be too difficult. Just another linear() function perhaps. The math is all there already...

Mylenium

Here's something that works:

rectSize = content("Rectangle 1").content("Rectangle Path 1").size;
rectRound=content("Rectangle 1").content("Rectangle Path 1").roundness;
rectStart=content("Rectangle 1").content("Trim Paths 1").start;
rectEnd=content("Rectangle 1").content("Trim Paths 1").end;

rectLen= rectSize[0] * 2 + rectSize[1] * 2+Math.PI*rectRound*2;

realStart=linear(rectStart,0,100,0,rectLen);
realEnd=linear(rectEnd,100,0,rectLen,0);

realDiff=realEnd-realStart;

value+rectLen/realDiff

There's still some slight creep due to precision issues the larger the rectangle and the rounded corner radius gets, but this definitely should do the trick.

Mylenium

All you have to do is divide the size of the rectangle by 360 and add that value to the offset. The expression looks like this:

var rectSize = content("Rectangle 1").content("Rectangle Path 1).size;
ofst = 360 / (rectSize[0] * 2 + rectSize[1] * 2);
value + ofst

Then set your roundness, start and end values, and adjust the offset to position the gap where you want it. The simple ratio generated by the expression on Offset keeps things lined up. That is by far the simplest solution.

Hint: Holding down the Cmnd/Ctrl key when dragging the offset value will divide the value by 10 and make fine-tuning the position of the gap easier. 

EDIT: Come to think of it, the roundness throws a kink in the expression. As it is, the expression keeps the gap the same size and allows you to position it anywhere. You will have to compensate for the roundness and make sure that it is limited to half the maximum width of the rectangle. I'll have to open up After Effects and fiddle with the expression for a while. I'll bet Dan Ebberts has the solution.