Let's find those square numbers, in whose decimal representation every digit is the same.
We denote by $U_{n}$ the $n$-digit number all of whose digits are 1. If $A$ has every digit as $a$, then $A=a \cdot U_{n}$ (for some $n$). The last digit of a square number in decimal notation can be $0, 1, 4, 5, 6$, and 9, so these are also the possible values for $a$. In the prime factorization of a square number, every prime number appears with an even exponent; since $U_{n}$ is not divisible by 2 or 5, $a$ cannot be 5 or 6. Therefore, $a$ must be a square number, and thus $U_{n}=\frac{A}{a}$ is also a square number. For $n=1$, $U_{1}=1$ is a square number, while for $n>1$, $U_{n}=100 U_{n-2}+11=4\left(25 U_{n-2}+2\right)+3$ cannot be a square number, since the square of an odd number $(2 k+1)^{2}=4\left(k^{2}+k\right)+1$. Therefore, the solutions to our problem are only the single-digit square numbers: $0, 1, 4$, and 9.