For example $18$ has a sum of digits equal to $1+8=9$, and when multiplied by any of those given numbers the resulting numbers sum of digits is still $9$.
I've realised that every number which has the sum of its digits equal to $9$ has this property that no matter what number you multiply it by you always preserve its sum of digits, but I don't know why only these numbers have this property
On
Every integer, whose digit sum is divisible is divisible by $9$, is divisible by $9$. However, although multiplying retains divisibility, the digit sums will most often be different.
As advised by @almagest, start by multiplying by $2$ to find the results $$18,27,36,45,54,63,72,81,90,99$$
Then multiply the above by $9$ to eliminate $54,63,72,81$
Next, use $8$ to eliminate $36$
Then $7$ to eliminate $27$
This leaves $18,45,90,99$ to be checked against multipliers $3,4,5,6$. There should be no further exclusions.
Your answer is four, which are $18,45,90,99$
Please let me know if you need more detail.
A number is divisible by 9 iff its digit sum is divisible by 9. So any number $n$ not a multiple of 9 will have a different digit sum from $9n$. That leaves us looking at: 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.
It is now a question of checking. $27\times7= 189$; $36\times 8=288$; $54\times 7=378$; $63\times 9=567$; $72\times 8=576$; $81\times 6=486$. So all those fail, leaving 18,45,90,99.
Answer: 4.