Suppose we have a 2 digit number $x$. We can write is in terms of its digits $a$ and $b$. When we attempt to square this number, we get an interesting result.

$$x^2=(10a+b)^2=100a^2+10(2ab)+b^2$$

We can also flip the digits (I'll use $\bar x$ to indicate this) and then square.

$$\bar x^2=(10b+a)^2=100b^2+10(2ab)+a^2$$

This result isn't very useful on its own, but if $a^2$, $b^2$, and $2ab$ are all less than $10$, then the three terms above are the three digits of $x^2$ and $\bar x^2$ respectively. It is clear from that that reversing the digits of $x$ reverses the digits of $x^2$ provided it meets those requirements. (If we switch $a$ and $b$, the first and last terms switch while the middle term is unchanged.)

Note that $10$, $11$, $12$, and $13$ (as well as $20$, $21$, $22$, $30$, and $31$) all satisfy the same condition $a^2,b^2,2ab<10$ and thus have the property you describe.

The same argument can be used for addition, since, if no digit is greater than $4$, we can add the digits individually. Rearranging the digits of such a number will apply the same rearrangement to its sum with itself.

We can play the same game with three digit numbers, but the restrictions are even greater:

$$x^2=(100a+10b+c)^2=10000a^2+1000(2ab)+100(b^2+2ac)+10(2bc)+c^2$$

If each factor multiplying a power of ten is less than ten, we have the same property. This gives us a few numbers, which you can verify all have the property.

$$100,101,102,103,110,111,112,113,120,121,122,130,200,201,202,210,211,212,220,300,301,310,311$$

Note that we have not shown that these conditions give you all the numbers for which $\bar x^2=\bar{x^2}$, though I haven't been able to find a counterexample. This argument nonetheless applies to all of the numbers you provided.