The number of the group homomorphsim , $\phi : S_3 \to S_3$ Surely, $S_3 = \langle(1,2), (1,2,3)\rangle$ 
And Put $\alpha = (1,2)$ and $\beta = (1,2,3)$ for homomorphism, $\phi : S_3 \to S_3$
Say $\phi(\alpha) =f(\in S_3)$ and $\phi(\beta) =r(\in S_3)$
Then, $\vert f \vert $ and $\vert r \vert$ should be divisor of the $2$ and $3$ respectively. 
Hence there are 4 cases like the below.
1) $\vert f \vert  =  1$ and $\vert r \vert  =  1$ (The number of the $\phi$ is $1$)
2) $\vert f \vert  =  1$ and $\vert r \vert  =  3$  (The number of the $\phi$ is $2$)
3) $\vert f \vert  =  2$ and $\vert r \vert  =  1$(The number of the $\phi$ is $3$)
4) $\vert f \vert  =  2$ and $\vert r \vert  =  3$(The number of the $\phi$ is $6$)
So my answer is $12$.
But the answer of that was $10$ not $12$. 
Which point I was wrong? Any advice would be appreciated. 

When classifying homomorphisms of a group, generally one cannot prescribe the orders of images of generators independently.
Indeed: If $\operatorname{ord}(\phi((12))) = 1$, then $\phi((12)) = 1$ (the identity permutation). But in a symmetric group any conjugacy class consists precisely of all permutations of a given cycle type, so there is some element $g \in S_3$ such that $$g (12) g^{-1} = (23) .$$ (I encourage you to verify this claim in this case by finding such an element $g$ explicitly.)

But since $\phi$ is a homomorphism, $$\phi((23)) = \phi(g (12) g^{-1}) = \phi(g) \phi((12)) \phi(g^{-1}) = \phi(g) \phi(g)^{-1} = 1$$ and so $$\phi((123)) = \phi((12)(23)) = \phi((12)) \phi((23)) = 1 \cdot 1 = 1.$$ In summary, the condition $\phi((12)) = 1$ in fact forces $\phi$ to be trivial, so there are no homomorphisms $S_3 \to S_3$ satisfying case (2). The remaining cases add up to $10$ distinct homomorphisms, but our observation in this answer shows that more is required to verify that all all of the remaining cases really are realizable.

Remark More generally, if a homomorphism $\psi : S_n \to S_n$ maps any transposition to $1$, then since all transpositions are conjugate, the above argument shows that $\psi$ maps all transpositions to $1$. But $S_n$ is generated by its transpositions, so any such $\phi$ is actually the trivial homomorphism.
This gives another illustration of the first statement in this answer. Another generating set for $S_3$ is $\{(12), (13)\}$, but since both of these generators have the same cycle type, their images under a homomorphism $S_3 \to S_3$ necessarily coincide.
