I think I have a general answer. First, the results:
All two-digit numbers that satisfy $r(n^2) = r(n)^2$, as found by Ankoganit, are:
11, 12, 13, 22, and their inverses.
All three-digit numbers are:
101, 102, 103, 111, 112, 113, 121, 122, 202, 212, and their inverses.
All four-digit numbers are:
1001, 1002, 1003, 1011, 1012, 1013, 1021, 1022, 1031, 1102, 1103, 1111, 1112, 1113, 1121, 1122, 1202, 1212, 2002, 2012, 2022, and their inverses.
(Larger numbers are listed at the end.)
The key to the solution is...
...we need to avoid carry, which is the only asymmetrical aspect here.
To understand this, it is very useful to think of squaring a number in terms of long multiplication. If we take, for example, a four-digit number, $n = 1000d_4 + 100d_3 + 10d_2 + d_1$, we have for $n^2$:
$\begin{array}{cccccccc}
& & & & d_4 & d_3 & d_2 & d_1 \\
\times & & & & d_4 & d_3 & d_2 & d_1 \\
\hline
& & & & d_1 d_4 & d_1 d_3 & d_1 d_2 & d_1^2 \\
+ & & & d_2 d_4 & d_2 d_3 & d_2^2 & d_2 d_1 \\
+ & & d_3 d_4 & d_3^2 & d_3 d_2 & d_3 d_1 \\
+ & d_4^2 & d_4 d_3 & d_4 d_2 & d_4 d_1 \\
\hline
& D_7 & D_6 & D_5 & D_4 & D_3 & D_2 & D_1
\end{array}$
(In case you're not familiar with long multiplication: each line in the middle section represents the multiplication of one digit by all the others. Then we sum each column to get $D_1, ..., D_7$, and these constitute the result, $n^2 = 10^6 D_7 + 10^5 D_6 + ... + D_1$.)
As long as all $D_i$'s are single digits (smaller or equal to 9) it doesn't matter if we reverse the digits before or after the square (notice the symmetric structure of the middle section). But once a certain $D_i$ is larger than 9, we get carry to the next digit. If, for example, $D_4 = 2(d_1 d_4 + d_2 d_3) > 9$, it would affect digit 5 in $n^2$ and thus digit 3 in $r(n^2)$. But the carry will be from digit 4 to 5 for $r(n)^2$ and it will stay as digit 5. And so we wouldn't have $r(n^2) = r(n)^2$.
A simple example is $n=2112$: We have $D_4 = 2(d_1 d_4 + d_2 d_3) = 2(4 + 1) = 10$ and so we get carry of 1. $r(2112^2) = 4450644$ (with carry 'to the right') and $r(2122)^2 = 4460544$ (with carry 'to the left'), and they are not equal.
So we want to avoid carry. But what causes carry? Anything that causes $D_i > 9$. First of all, as you can see above, all digits appear as $d_i^2$ in one of the columns. So any digit which has $d_i^2 > 9$ causes carry and is thus forbidden. So all digits must be 1, 2 or 3, as other people already said.
Second, all pairs of digits appear twice as $d_i d_j$ in one of the columns. So any pair of digits which has $2 d_i d_j > 9$ causes carry and is thus forbidden. So we cannot have both 2 and 3 in $n$ (as $2 \cdot 2 \cdot 3 > 9$), nor can we have two 3's (as $2 \cdot 3 \cdot 3 > 9$).
These two rules are sufficient to construct all possible two-digit $n$'s, listed above.
Note that we only had to look at the columns of $D_1$ and $D_2$ to come up with forbidden two-digit $n$'s. For three-digit $n$'s we also look at the column of $D_3$. It tells us we need to avoid a sequence of three consecutive digits, $d_1$, $d_2$ and $d_3$, which have $d_2^2 + 2 d_1 d_3 > 9$. These are (given the former rules) the sequence 2-2-2 and the sequence 1-3-1. So we can construct all three-digit $n$'s, listed above.
For a four-digit $n$, we still have to avoid the three-digit sequences above (whether they come as $d_1, d_2, d_3$ or $d_2, d_3, d_4$) and we look at $D_4$ to get the forbidden four-digit sequences: 1221, 2112, 2122 and 2212 (which all have $2(d_1 d_4 + d_2 d_3) > 9$). We construct all four-digit $n$'s, listed above.
And so we go on and on. Any $k$-digit forbidden sequence is also forbidden for $n$'s with more digits.
Note that when we get to 5 digits, three-digit forbidden sequences don't have to be consecutive. They can come as $d_1$, $d_3$ and $d_5$, and appear as $d_3^2 + 2 d_1 d_5$ in column 5. More generally, every forbidden sequence is forbidden whenever the digits are equidistant (not necessarily at distance 1), as they all contribute to the same column. We take this for granted when it's a pair of digits, but it is true generally.
BTW, the lack of carry means all possible $n^2$'s (and $r(n)^2$'s) are of an odd number of digits. Ankoganit showed that no four-digit $n^2$'s are possible, but in fact this is true for any even number of digits.
Different base: These rules are easily converted to any base, since the key is still avoiding carry. For base-$D$ (of $D$ 'digits'), we have $r(n^2) = r(n)^2$ iff:
- all digits of $n$ satisfy $d_i^2 < D$,
- all pairs of digits of $n$ satisfy $2 d_i d_j < D$,
- all 'equidistant' triplets of digits of $n$ satisfy $d_j^2 + 2 d_i d_k < D$ (with $d_j$ in between $d_i$ and $d_k$),
- and so on.
So, for example, in base-17, the possible digits are $0, 1, 2, 3,$ and $4$. Then, the pairs $(3, 3)$, $(3, 4)$ and $(4, 4)$ are forbidden. Also, the triplets $(2, 1, 4)$, $(2, 2, 4)$, $(2, 3, 2)$, $(1, 4, 1)$, $(1, 4, 2)$, $(2, 4, 1)$ and $(2, 4, 2)$ are forbidden, and so on.
In bits (base-2), both 0 and 1 are allowed. But the pair $(1, 1)$ is not, so all $n$'s must have a single 1 at most. So the only $n$'s that work are powers of 2 (which all have $r(n^2) = 1$). But, in fact, $r$ is not defined when the first or the last digit is 0, so strictly speaking, the only possible $n$ in base-2 is 1...
For whoever is interested, these are all (base-10) $n$'s of five to seven digits that satisfy $r(n)^2 = r(n^2)$:
$10001, 10002, 10003, 10011, 10012, 10013, 10021, 10022, 10031, 10101, 10102, 10103, 10111, 10112, 10113, 10121, 10122, 10201, 10202, 10211, 10212, 10221, 11002, 11003, 11011, 11012, 11013, 11021, 11022, 11031, 11102, 11103, 11111, 11112, 11113, 11121, 11122, 11202, 11211, 12002, 12012, 12102, 12202, 20002, 20012, 20022, 20102, 20112, 20122, 100001, 100002, 100003, 100011, 100012, 100013, 100021, 100022, 100031, 100101, 100102, 100103, 100111, 100112, 100113, 100121, 100122, 100201, 100202, 100211, 100212, 100221, 100301, 100311, 101002, 101003, 101011, 101012, 101013, 101021, 101022, 101031, 101101, 101102, 101103, 101111, 101112, 101113, 101121, 101122, 101201, 101202, 101211, 101212, 101301, 102002, 102011, 102012, 102021, 102022, 102102, 102111, 102121, 110002, 110003, 110011, 110012, 110013, 110021, 110022, 110031, 110102, 110103, 110111, 110112, 110113, 110121, 110122, 110202, 110211, 110212, 110221, 111002, 111003, 111012, 111013, 111021, 111022, 111031, 111102, 111103, 111111, 111112, 111121, 111202, 111211, 112002, 112012, 112102, 120002, 120012, 120102, 120112, 121002, 121102, 122002, 200002, 200012, 200022, 200102, 200112, 200122, 200202, 200212, 201012, 201022, 202012, 1000001, 1000002, 1000003, 1000011, 1000012, 1000013, 1000021, 1000022, 1000031, 1000101, 1000102, 1000103, 1000111, 1000112, 1000113, 1000121, 1000122, 1000201, 1000202, 1000211, 1000212, 1000221, 1000301, 1000311, 1001001, 1001002, 1001003, 1001011, 1001012, 1001013, 1001021, 1001022, 1001031, 1001101, 1001102, 1001103, 1001111, 1001112, 1001113, 1001121, 1001122, 1001201, 1001202, 1001211, 1001212, 1001301, 1002001, 1002002, 1002011, 1002012, 1002021, 1002022, 1002101, 1002102, 1002111, 1002121, 1002201, 1002202, 1002211, 1010002, 1010003, 1010011, 1010012, 1010013, 1010021, 1010022, 1010031, 1010101, 1010102, 1010103, 1010111, 1010112, 1010113, 1010121, 1010122, 1010201, 1010202, 1010211, 1010212, 1010221, 1011002, 1011003, 1011011, 1011012, 1011013, 1011021, 1011022, 1011031, 1011101, 1011102, 1011103, 1011111, 1011112, 1011113, 1011121, 1011122, 1011201, 1011202, 1011211, 1012002, 1012011, 1012012, 1012101, 1012111, 1020002, 1020011, 1020012, 1020021, 1020022, 1020102, 1020111, 1020112, 1020121, 1020122, 1021002, 1021011, 1021021, 1021102, 1021111, 1022002, 1030011, 1031011, 1100002, 1100003, 1100011, 1100012, 1100013, 1100021, 1100022, 1100031, 1100102, 1100103, 1100111, 1100112, 1100113, 1100121, 1100122, 1100202, 1100211, 1100212, 1100221, 1100311, 1101002, 1101003, 1101011, 1101012, 1101013, 1101021, 1101022, 1101031, 1101102, 1101103, 1101111, 1101112, 1101113, 1101121, 1101122, 1101202, 1101211, 1101212, 1102002, 1102011, 1102102, 1102111, 1110002, 1110003, 1110012, 1110013, 1110021, 1110022, 1110031, 1110102, 1110103, 1110111, 1110112, 1110121, 1110202, 1110211, 1111002, 1111003, 1111012, 1111013, 1111021, 1111022, 1111031, 1111102, 1111103, 1111111, 1111112, 1111121, 1111202, 1111211, 1112002, 1120002, 1120012, 1120102, 1121002, 1121102, 1122002, 1200002, 1200012, 1200102, 1200112, 1200202, 1201002, 1201012, 1202002, 1210002, 1210102, 1210202, 1211002, 1212002, 1220002, 1220102, 2000002, 2000012, 2000022, 2000102, 2000112, 2000122, 2000202, 2000212, 2001002, 2001012, 2001022, 2001102, 2001112, 2001122, 2001202, 2001212, 2010012, 2010022, 2011012, 2020012, 2020022,$ and their inverses.
11, 22, 33, ..., 101, 111, 121, ...
$\endgroup$