Group Isomorphism Problem Let $S=\{(k,x) | k \in \mathbb{F}_5 \setminus \{0\}, x \in \mathbb{F}_5 \} $ be the group with binary operation $(k,x)*(l,y)=(kl,xl+y)$. Let $P$ be a Sylow 5-subgroup of $G=Sym(5)$. I am asked to show that the normaliser $N_G(P)$ is isomorphic to $S$.
In the previous parts of the question I was asked to find a normal subgroup of $S$ that is isomorphic to $\mathbb{F}_5$, which I found to be $T=\{(1,t) | t \in \mathbb{F}_5 \} $ . I was then asked to find the order of $N_G(P)$, and I found it to be 20, which is consistent with what is asked above.
It is clear that $T\cong P$ since they are both cyclic groups of order 5. I thought by showing  $N_S(T)=S$ (which is obvious since $T$ is normal in $S$), it implies that $N_G(P) \cong N_S(T) =S$. But it turns out it's not enough (I was pretty naive to think that would prove it lol). Now I am all out of ideas.
Any ideas?  
 A: Let $P$ be generated by the cycle $\alpha:=(12345)$. What are the permutations in $ N_G(P)$ be a permutation?
E.g. if $(\sigma(1),\sigma(2),\sigma(3),\sigma(4),\sigma(5))=(1,3,5,2,4)$ then the conjugate permutation satisfies 
$\sigma\alpha\sigma^{-1}=(13524)=\alpha^2$, so this $\sigma=(2354)$ is in $N_G(P)$. Now, as this has order $4$, and $\alpha\sigma=\alpha\sigma^{-3}=\sigma^{-3}\sigma^3\alpha\sigma^{-3}=\sigma^{-3}\left((\alpha^2)^2\right)^2 =\sigma\alpha^3$, 
 we could try to extend the mapping
$$\alpha\mapsto(1,1)\quad\quad \sigma\mapsto(3,0)\,. $$
First, observe that, as $\alpha^x\sigma=\sigma\alpha^{3x}$, we have
$$\sigma^i\alpha^x\sigma^j\alpha^y=\sigma^{i+j}\alpha^{3^jx+y}\,.$$
So that the mapping $(3^i,x)\mapsto \sigma^i\alpha^x$ seems indeed a homomorphism. It also proves that $\{\sigma^j\alpha^n\,\mid\,j=0...3,\ n=0...4\}$ is a subgroup.
All we are left to prove is that $N_G(P)$ has no more elements.
Now if $\vartheta\in N_G(P)$, then $\vartheta\alpha\vartheta^{-1}=\alpha^k$ for some $k$. Comparing this to $\alpha^k=\sigma^j\alpha\sigma^{-j}$ for some $j$, we get
that $\vartheta\sigma^{-j}$ commutes with $\alpha$, i.e. $=\alpha^n$ for some $n$, but then $\vartheta=\sigma^j\alpha^n$.
