What did I wrong in this question?(Group homomorphism and primitive roots) This is the question that don't  have any idea Which point I was wrong
Q)
Let group homomorphism $f : Z_{50}^* \to  Z_{50}^*  $ $by$ $f(3) = 31$
(Here, the $Z_{50}^* = \{a \in Z_{50} \vert gcd(a,50)=1\}$ )
$3$ is a primitive root for $mod$ $50$
Find all the element of the $A =  \{ x \in Z_{50}^* \vert f(x)=9\}$

My attempt) $31 = 81 = 3^4$ $(mod50)$ So, $f(3) = 3^4$
Plus Owing to the $3$ is a primitive roots of the $Z_{50}^*$, $3$ is a generator of the group $Z_{50}^*$
Then All we have to do is just find the $3^a s.t.$ $f(3^a) = 3^{4a} =3^2(mod50)$
Hence Find the $a$ satisfying $4a = 2(mod \phi(50))$
(Here the $a \in \{x \vert 1 \leq x <50, gcd(x,50)=1\}$)
But $\phi(50)$ = $20$, There aren't exist the "$a$ ". (I.E. $A = \phi$)
p.s.)
But the someone who gave me this question said the answer is $A = \{9,13,37,41\}$
I totally couldn't understand Which point I was wrong.
Please help me. Thanks.
Additional post) 
Here is that person's solution who claiming the $A = \{9,13,37,41\}$
He might be the suggesting the incorrect answer, Surely there are two possibility that Should be incorrect either mine or his.
For the integer set $Z$, since $3^4 = 31$, 
$imf=\{f(3^a) \vert a \in Z\} = \{(3^4)^a \vert a \in Z\}= <3^4>$
Then $\vert imf \vert  =5$ and $\vert Z_{50}^* / kerf \vert = \vert imf \vert = 5$
Hence $\vert kerf \vert =4$
Also, $f(-7) = f(3^5) = f(3)^5 = 3^{20} =1$
Plus, $f(-1) = f(3^{10} ) = f(3)^{10} = 3^{40} =1$ 
Therefore $\{-1, -7\} \in kerf$ So, $kerf = \{1,-1,7,-7\} $
We can conclude the $A = f^{-1}({9}) = 9kerf = \{9,-9,63,-63\} = \{9,13,37,41\}$
 A: Since you have identified $3^4=31$, have been given that $3$ is a primitive root and know that the order of the group is $20$ you know that $31$ will generate a group of order $5$ as the image of the homomorphism, consisting of the fourth powers of elements in the original group.
$9$ is not a fourth power, so isn't in the image, so the pre-image is empty as you have concluded.
A: I cannot see where you are wrong.  Since $13=3^{17} \bmod 50$, you have $f(13)=f(3^{17})=31^{17}=11 \neq 9$.  Moreover, $9=3^2 \bmod 50$ and so $f(9)=31^2 =11 \neq 9$.   Can you check the other proposed solutions?  Why must the other person be right?
(Incidentally, $f(37)=11$ and $f(41)=11$, too, so at least the other person is wrong.)
After your edits, your colleague is wrong.  He is using the fact that the kernel has the same size as any pre-image of a point, which is correct AS LONG AS THE PRE-IMAGE IS NON-EMPTY.  The correct statement of that theorem goes as follows:
Let $\phi: G \to K$ be a group homomorphism.  Then the set $\phi^{-1}(\phi(a))$ is equal to the coset $a \mathrm{ker}(\phi)$.  In other words, if $\phi(a)=b$ then $\phi^{-1}(b) = a \mathrm{ker}(\phi)$.  
Your colleague is wrong, because there is no such $a$ for $b=9$.
For a concrete example of this mistake, the inclusion homomorphism $i: \mathbb{Z} \to \mathbb{R}$ has kernel with one element, but not every pre-image has one element:  what's the pre-image of $\{0.5\}$?
