This is a follow-up question to Existence/uniqueness of the Teichmuller map for p-adic integers
Let $Z_p$ denote the p-adic integers. Let $π:Z_p \to \mathbb{Z}/p\mathbb{Z}$ by $π(a_0+a_1p+a_2p^2+...)=a_0$ where, of course, each $a_i \in \{0,1,...,p-1\}$.
It turns out that there exists some unique function $T:\mathbb{Z}/p\mathbb{Z}\to Z_p$ satisfying the following two properties:
$\forall x \in \mathbb{Z}/p\mathbb{Z}[π(T(x))=x]$
$\forall x,y \in \mathbb{Z}/p\mathbb{Z}[T(x)^p=T(x)]$
Moreover, it also turns out that $\forall x,y \in \mathbb{Z}/p \mathbb{Z}[T(xy)=T(x)T(y)$]
Now, suppose that $χ: \mathbb{Z}/p \mathbb{Z} \to Z_p$ is an arbitrary function satisfying $\forall x,y \in \mathbb{Z}/p \mathbb{Z}[χ(xy)=χ(x)χ(y)$]
According to a proof I'm following, the author seems to take for granted that $χ(x) \equiv T(x)^j$ for some $j$. But I don't see why that's true. It's easy to see why any power of $T$ must be a multiplicative character, but the converse isn't obvious to me.
My attempt so far
I see that $T^p=T$, so there can be at most p distinct powers of $T$. If we can prove the following, we'll be done:
$T^0,T^1,...,T^{p-1}$ are distinct
There exist no more than $p$ distinct possible multiplicative characters $χ$
Assume $p$ odd. (For $p=2$, slight changes apply.)
Notice that such a $\chi$ must in particular induce a group homomorphism $(\Bbb Z/p\Bbb Z)^\times \rightarrow \mathbb Z_p^\times$. Now the group on the left is cyclic of order $p-1$. So if you fix any of its generators, call it $x$, then $\chi$ is uniquely determined by where it sends that generator.
I claim there are $p-1$ possible choices for $\chi(x)$, i.e. elements of $\mathbb Z_p^\times$ whose ($p-1$)-th power is $1 \in \mathbb Z_p$ (can you see why, and which elements I mean?), and each of them can be written as a power of $T(x)$ (why?).
So $\chi(x) = T(x)^j$. But then $\chi(y) =T(y)^j$ for all $y \in (\Bbb Z/p\Bbb Z)^\times$ (why?).