A seemingly strange function may be quite simple instead

1. Suppose $\displaystyle\large n=\overline{x_{k}\cdots x_{1}x_{0}}$ ,here $x_{i}\in \{1,2,3,4,5,6,7,8,9\}$ for $i=0,1,\cdots,k$ .

Then $\displaystyle\large 10^{k}<n<10^{k+1}$ ,which implies $\displaystyle\large k<lgn<k+1$ .Thus $\displaystyle\large \lfloor lgn \rfloor = k$ .

Now easy to see $\displaystyle\large \Big\lfloor\frac{n}{10^{\lfloor lgn \rfloor}}\Big\rfloor =x_{k}$ .

So $\displaystyle\large S(n)=x_{k}+10\big(\overline{x_{k}\cdots x_{1}x_{0}}-\overline{x_{k}\underbrace{0\cdots 00}_{k-1}}\big)=\overline{x_{k-1}\cdots x_{1}x_{0}x_{k}}$ ,

and $\displaystyle\large S(S(n))=\overline{x_{k-2}\cdots x_{1}x_{0}x_{k-1}x_{k}}$ .

2. Apply the result above. If n has the digit $0$ , then n cannot equal to $S(S(n))$ . (You can check it in Sir. Mark Hennings's comment below.)

In the following statement, $a,b,c,d$ are positive integers less than 10, $|A|$ stands for the number of elements in set $A$ .

$S(S(\overline{a}))=\overline{a}$ $\longrightarrow$ $A_{1}=$ $\{1,2,3,4,5,6,7,8,9\}$ , $|A_{1}|=9$

$S(S(\overline{ab}))=\overline{ab}$ $\longrightarrow$ $A_{2}=$ $\{11,12,\cdots ,19; 21,22,\cdots ,29; \cdots ;91,92,\cdots ,99\}$ , $|A_{2}|=81$

$S(S(\overline{abc}))=\overline{cab}$ $\longrightarrow$ $A_{3}=$ $\{111,222,\cdots ,999\}$ , $|A_{3}|=9$

$S(S(\overline{abcd}))=\overline{cdab}$ $\longrightarrow$ $A_{4}=$ $\{1111,1212,1313,\cdots ,1919; 2121,2222,2323,\cdots ,2929; \cdots ; 4141,4242,\cdots ,4949\}$ , $|A_{4}|=36$

So the answer is $\displaystyle\sum^{4}_{i=1}{|Ai|}=\boxed{135}.$