Let's find a 6-digit number that is a perfect square and remains the same when the order of its digits is reversed.
Let $N$ be the sought number and $N^{\prime}$ the reversed number; further, let $N=p^{2}$. Since every square ends in $0,1,4,5,6$ or 9, the first digit of $N$ must also be one of these numbers, otherwise $N^{\prime}$ could not be a perfect square.

Since

$$
N=a 10^{5}+b 10^{4}+c 10^{3}+d 10^{2}+e 10+f
$$

and

$$
\begin{gathered}
N^{\prime}=f 10^{5}+e 10^{4}+d 10^{3}+c 10^{2}+b 10+a \\
N+N^{\prime}=(a+f) 10^{5}+(b+e) 10^{4}+(c+d) 10^{3}+ \\
+(d+c) 10^{2}+(e+b) 10+f+a
\end{gathered}
$$

However, in this number, the sum of the digits in the even positions is equal to the sum of the digits in the odd positions, that is,

$$
(a+f)+(c+d)+(e+b)=(b+e)+d+c)+(f+a)
$$

thus $N+N^{\prime}=k \cdot 11$.

However, the sum of two perfect squares is divisible by 11 only if each is individually divisible by 11, and thus

$$
N=m \cdot 11
$$

and

$$
p=n \cdot 11
$$

Since $N$ is a six-digit number,

$$
\begin{aligned}
& 10^{5}<N<10^{6} \\
& 316<p<1000
\end{aligned}
$$

Furthermore, since $N$ starts with 1, 4, 5, 6, or 9, $N$ can only lie within the following ranges:

100000 and 200000

400000 and 700000

900000 and 1000000.

$$
316<p<448
$$

or

$$
632<p<837
$$

or finally

$$
918<p<1000
$$

We see, therefore, that $p$ is a multiple of 11 that lies within the above ranges.

To facilitate further searching, note that if the first digit of $N$ is 5, the last digit of $N^{\prime}$ is also 5; and since every perfect square ending in 5 has a 2 in the tens place, the second digit of $N$ is also 2, and thus

$$
520000<N<530000
$$

or

$$
721<p<729
$$

Since 726 is the only multiple of 11 within these bounds and its square does not meet the problem's requirements, we can state that $N$ does not lie between 500000 and 600000, and $p$ cannot appear between 708 and 774, and it cannot end in 5. Furthermore, note that if the last digit of a perfect square is 1, 4, or 9, the tens place digit is even, and if the last digit is 6, the second-to-last digit is odd. If, therefore, the first digit of $N$ is 1, 4, or 9, the second digit is even, while if the first digit is 6, the second digit is odd.

Indeed, the squares of numbers ending in 1, 4, and 9 are of the form:

$$
10 q+1, \quad 10 q \pm 2, \quad 10 q \pm 3
$$

and these are

$$
100 q^{2}+10 \cdot 2 q+1, \quad 100 q^{2} \pm 10 \cdot 4 q + 4, \quad 100 q^{2} \pm 10 \cdot 6 q+9
$$

while those ending in 6 are of the form

$$
10 q \pm 4
$$

whose squares are

$$
100^{2} \pm 10 \cdot 8 q+16=100 q^{2}+10( \pm 8 q+1)+6
$$

which confirms our statements above.

After all this, only the following multiples of 11 can be considered:

$$
319, \quad 330, \quad 352, \quad 407, \quad 429, \quad 638, \quad 649, \quad 682, \quad 693, \quad 836, \quad 990
$$

of which

$$
330, \quad 836 \text { and } 990
$$

satisfy the problem. Their squares are indeed

$$
\text { 108900, } 698896, \quad 980100
$$

and their reversals also yield perfect squares.