If _ _ _ _ _ $9 \times 4 = 9 $_ _ _ _ _, what's the former number? 
_ _ _ _ _ $9 \times 4 = 9 $_ _ _ _ _ 
I have a 6-digit positive integer whose last digit is 9. If I move
  this last digit directly to the front to form another 6-digit integer,
  then the original number is quadrupled.
What is the original 6-digit number?

Let $\large{\overline {ABCDE9}\times4=\overline {9FGHIJ}}$
So far, using trial and error, I could figure out the list of possibilities for
$A \rightarrow2$
$B \rightarrow3,4$
$E \rightarrow3, 8,1 ,6,4,9$
$J \rightarrow6$
$I \rightarrow5,7,9$
I'm stuck at this point. I have no clue how to narrow down the set of possibilities.

EDIT
How can I possibly assume that the digits which fill the blanks of one side are identical to those of the other side?

EDIT$^{2}$
I had actually misinterpreted the question. I thought that $ABCDE$ and $FGHIJ$ are distinct. 
I started reworking on the problem and came up with two different solutions, all of which are already covered by answers (of Simply Beautiful Art and Micheal Burr). 
 A: Hint:
$$\underbrace{x=abcde9}_{\text{we want to solve for this}}$$
$$\underbrace{900000+abcde=4x}_{\text{this is moving the 9 to the front}}$$
$$\underbrace{abcde0=x-9}_{\text{this is dropping the 9 from the end}}$$
$$\underbrace{abcde=\frac1{10}abcde0}_{\text{this is moving the digits over}}$$
Thus,
$$4x=900000+\frac1{10}(x-9)$$
Solving this gives

 $x=230769$

A: I am interpreting this question in a different way than the OP.  The question is about the operation being performed.  The question states that you start with a $6$ digit number which ends in a $9$, say 
$$
123459
$$
and move the last digit (the $9$) to the front, you have a new $6$ digit number, in this case
$$
912345.
$$
Observe that the other $5$ digits are unchanged, just their position changes.  This matches the interpretation of @SimplyBeautifulArt.  In the OP's question, the digits (other than $9$) can be anything, and there will not be a unique answer unless the structure between the other digits is used.
Now, this example doesn't satisfy the given conditions because it doesn't have the multiplication by $4$ property.  Now, this problem can be solved directly.  You know that 
$$
4\cdot abcde9=9abcde
$$
Since the units digit of the product on the left is $6$, $e=6$.  Therefore, we now have 
$$
4\cdot abcd69=9abcd6.
$$
The $10$s digit on the left is $7$, so $d=7$.  Hence, we have 
$$
4\cdot abc769=9abc76.
$$
Continuing wit the $100$s digit, we get that $c=0$.  Therefore, we now have 
$$
4\cdot ab0769=9ab076.
$$
For the $1,000$s digit, we get that $b=3$, so we now have 
$$
4\cdot a30769=9a3076.
$$
Finally, the $10,000$s digit on the LHS is $2$, so $a=2$.  Thus, we have
$$
4\cdot 230769=923076.
$$
Therefore, the original number is $230769$.
A: Assuming $ABCDE$ is $abcde$ is the case:
$$4[(abcde)\cdot10+9]=9\cdot10^5+abcde$$
That is,
$$39(abcde)=9\cdot(10^5-4) => abcde=23076$$
So original is simply $230769$.

Note: Your method is too time consuming, and if only not possible to use equations, must you resort to trial and error.
A: It says that if you move the last digit, which is a 9, to the front, then the number gets quadrupled.
So, you have a 6-digit number ABCDE9, so if you move the 9 to the front, you get 9ABCDE. We are told that this new number 9ABCDE is 4 times the origial number, so you need to solve:
ABCDE9 x 4 = 9ABCDE
A: So, the original number is in the form of  $\overline{abcde9}$ and the second one is $\overline{9abcde}$. Therefore, you can "translate" it as:
$$
4\cdot\overline {abcde9}=\overline {9abcde}\tag1
$$
Now, let $x = \overline{abcde}$:
$$
\overline {abcde9} = 10x + 9
$$
$$ \overline {9abcde} = 9 \cdot 10^5 + x
$$
From (1):
$$
4\cdot(10x + 9) = 9 \cdot 10^5 + x
$$
$$
40 x + 36 = 9\cdot 10^5+x
$$
$$
39x = 899964 \implies x = 23076
$$
The original number is $\overline{x9}=230769$
A: $\overline{abcde9}= 10^5a + 10^4b + 10^3c + 10^2d + 10e + 9$
$\overline{9abcde}= 9\times10^5 + 10^4a + 10^3b + 10^2c + 10d + e$
Let $x=\overline{abcde9}$. Making $10e$ the subject of the top equation: 
10e = x - 100,000a - 10,000b - 1000c - 100d - 9 Divide both sides by 10.
e= 
e=  -10,000a - 1000b - 100c - 10d - 0.9
As you can see, plugging this value of e into the second equation above conveniently cancels out all the other variables.
{x}=900,000 + 10,000a + 1000b + 100c + 10d +  -10,000a - 1000b - 100c - 10d - 0.9
{x} = 899,999.1 + 
4x = 899,999.1 + 
x = 899,999.1
x = 899,999.1 / 39 * 10
x = 230,769
