Suppose $n$ has $k$ digits, hence $10^{k-1}\le n<10^k$. Then $10^{2k-2}\le n^2<10^{2k}$ and $n^2$ has at most $2k$ digits. Hence $10^{k-1}\le n =s(n^2)\le 18k \implies k\le 2 \implies n\le 36$. Also $n=s(n^2)\equiv n^2 (9)\implies n(n-1)\equiv 0(9)\implies n\equiv 0,1(9)$, since $n,n-1$ cannot be both divisible by 3. Checking $n=9,18,27,36,1,10,19,28$ gives the solutions $n=1,9$.