Solution:

We first count the number of substrings that one could pick to reverse to yield a new substring. If we insert two dividers into the sequence of $50$ digits, each arrangement of $2$ dividers among the $52$ total objects specifies a substring that is contained between the two dividers, for a total of $\binom{52}{2}$ substrings. Next, we account for overcounting. Every substring of length $0$ or $1$ will give the identity string when reversed, so we are overcounting here by $51+50-1=100$ substrings. Next, for any longer substring $s$ that starts and ends with the same digit, removing the digit from both ends results in a substring $s'$, such that reversing $s$ would give the same string as reversing $s'$. Therefore, we are overcounting by $10 \cdot \binom{5}{2}$ substrings. Our total number of strings similar to $S$ is therefore $\binom{52}{2}-100-10 \cdot \binom{5}{2}=1126$.