Don't use algebra to find the unknown

$\overline{ABCDE9} \times 4=\overline{9ABCDE}$

One thing we know for certain is that $A$ could be any integer from $0$ to $9$ . So, 4 times any digit cannot exceed $36$ which is $4\times9$ .

We can deduce the following:

• $A$ cannot be $0$ (otherwise the first number would turn a 5-digit integer).
• $4$ multiplied with $A$ should yield a single-digit integer (both the numbers have the same number of digits).
• $4A$ should be less than or equal to $9$ (but not equal to $0$ ).

If we choose to rely on the premises above, the set of possibilities for $A$ gets narrowed.

This leaves us with two possibilities ( $1$ or $2$ ):

$4 \times \boxed{1}=4$ ——— $(1)$

$4 \times \boxed{2}=8$ ——— $(2)$

The tens digit of $4B$ adds with either of the products above to make $9$ .

$(1)$ — $4+5=9$

$(2)$ — $8+1=9$

For option (1) to be valid, the tens digit of $4B$ needs to be $5$ . But since $36$ is the maximum possible value of $4B$ , the tens digit of $4B$ cannot possibly be more than $3$ . Thus, option (1) cannot be valid.

We are finally left with option (2). Here, the tens digit of $4B$ is $1$ (which is well within the range).

$\therefore A$ is $\boxed{2}$ .