Why complement of most negative number in octal does not turn out to be itself
First of all, if you are considering the $n$th digit of the number as being a sign digit, you can think of it as being either 0 or 1, no matter the value of $r$. In other words, it is always a bit (a  binary digit), and that's how you would represent it in hardware, should that be the case.


  $[1]73_{\text{8's comp}} = -5$
  
  $[1]73_{\text{8's comp}} + [0]06_{\text{8's comp}} = [0]01_{\text{8's comp}}$


The complement of a negative number represented in $r$'s complement will obviously produce the positive version of that same number. The most negative number doesn't have a positive of the same magnitude that fits into the given number of digits (a weakness of this form of representation). 

The number is still there, though, only unsigned:


  $[1]00_{\text{8's comp}} = -64_{10}$
  
  $100_8 = 64_{10}$