For $G/F$ a connected reductive group and $LG / k$ we made the following definition.
Definition 1. An affine pinning of $LG$ consists of a triple $(I, T, \phi)$ where
- $I \subseteq LG$ is an Iwahori,
- $T \subseteq I$ is a maximal torus,
- $\phi \colon I^+ \to \mathbb{G}_ a$ is a homomorphism, where $I^+$ is the pro-unipotent radical, where under the factorization $I^+ \twoheadrightarrow I^+ / [I^+, I^+] \cong \coprod_\alpha U_\alpha \to \mathbb{G}_ a$ the restriction to each $U_\alpha$ is an isomorphism.
We now want to look at $$ M_\phi = N_{LG}(I, T, \phi) $$ the normalizer of the pinning. Here, all $(I, T)$ are conjugate to each other, but not all $(I, T, \phi)$ are conjugate to each other.
Remark 2. Consider $F = k((\varpi))$ where $\mathbb{G}_ m^\mathrm{rot}$ acts on $F$, and hence on $LG$. If $G$ is almost simple, then all affine pinnings of $LG$ are conjugate by $LG \rtimes \mathbb{G}_ m^\mathrm{rot}$, but they are not conjugate by $LG$.
Proposition 3. There is a short exact sequence $$ 1 \to Z_G \to M_\phi \to \Omega = \tilde{W}/W_\mathrm{aff} \cong \pi_0(LG) \to 1. $$ Moreover, $M_\phi$ is commutative.
Recall we have the Moy–Prasad filtration $$ I \vartriangleright I^+ \vartriangleright I^{++} \vartriangleright \dotsb $$ where $I / I^+ = T$ and $I^+ / I^{++} = V_I \cong \operatorname{Lie} I^+ / \operatorname{Lie} I^{++}$. (This is a general fact about Moy–Prasad filtrations.)
We see that there is a short exact sequence $$ 1 \to T \to M = N_{LG}(I, T) \to \Omega \to 1. $$ Now we want to understand $M_\phi = N_M(\phi)$. It is not hard to see that $M_\phi \cap T = Z_G$; the $T$ acts on $\phi$ through each root, and for this to be trivial we need to be in the center $Z_G$.
Now we need to show surjectivity. For each $\omega \in \Omega$, there is a lift in $\dot{\omega} \in M$. Then for each $\Phi_\mathrm{aff}$ simple, we need the diagram $$ \begin{CD} U_\alpha @>{\phi_\alpha}>> \mathbb{G}_ a \br @V{\dot{\omega}}VV @| \br U_{\omega(\alpha)} @>{\phi_{\omega(\alpha)}}>> \mathbb{G}_ a \end{CD} $$ to commute. But right now it only commutes up to a factor $c_\alpha \in k^\times$. Here, if we change $\dot{\omega}$ with $t \dot{\omega}$ then we are changing $c_\alpha$ to $\omega(\alpha)(t) c_\alpha$. But this is overdetermined, so there is some extra information we need to check that we can make all $c_\alpha$ to be $1$.
For simplicity, assume that $G$ is almost simple. Then some number $\prod_i c_i^{n_i}$ is independent of the choice of $\dot{\omega}$. This invariant can also be thought of as the GIT quotient map $$ h \colon V_I^\ast \to V_I^\ast // T \cong \mathbb{A}^1; \quad (c_i) \mapsto \prod_i c_i^{n_i}. $$ So all we need is the following lemma. (By $M_\phi$ we mean the group scheme, so we don’t need to worry about rational points.)
Lemma 4. Because $M$ acts on $V_I^\ast$, the group $\Omega$ acts on $V_I^\ast // T$. The induced action of $\Omega$ on $V_I^\ast // T$ is trivial.
All of these are actually defined over $\mathbb{Z}$, so we may as well work with the field $k = \mathbb{C}$. In this case, we have the Killing form $\mathfrak{g}^\ast \cong \mathfrak{g}$.
In general, we have a periodic chain $$ \operatorname{Lie} I \supseteq \operatorname{Lie} I^+ \supseteq \dotsb \supseteq \varpi \operatorname{Lie} I $$ and if we dualize this we get $$ (\operatorname{Lie} I)^\ast \subseteq (\operatorname{Lie} I^+)^\ast \subseteq \dotsb \subseteq \varpi^{-1} (\operatorname{Lie} I)^\ast. $$ On the other hand, there is a Killing form $B \colon \mathfrak{g} \otimes \mathfrak{g} \to F$, which induces a perfect pairing $$ L\mathfrak{g} \otimes L\mathfrak{g} \to F \xrightarrow{\mathrm{res}} k, \quad \mathrm{res} \colon \sum a_i \varpi^i \mapsto a_{-1}, $$ in the sense that $(L\mathfrak{g})^\ast \cong L\mathfrak{g}$ where $\ast$ is the topological dual. Let us continue this next time.
Lemma 5. The group $M_\phi$ is abelian.
To show this, it is enough to show that the commutator pairing $\Omega \times \Omega \to Z_G$ is trivial. Again, we reduce to $k = \mathbb{C}$. In this case, we may regard $\phi$ as an element of $$ \phi \in LG = \mathfrak{g} \otimes k((\varpi)), \quad \phi = \varpi X_\theta + \sum_a X_a. $$ Here, $\phi \in \mathfrak{g} \otimes k[[\varpi]]$ and it is regular nilpotent modulo $\varpi$. It follows that it is regular in $\mathfrak{g} \otimes F$ as well, and therefore $$ L C_G(\phi) = C_{LG}(\phi) \supseteq M_\phi $$ is commutative.
Now if $Z(G)$ is a torus, we have an exact sequence $$ 1 \to Z(G)(k) \to M_\phi(k) \to \Omega \to 1. $$ Recall that there are actions $$ {}_ \chi \mathcal{H}_ \chi \curvearrowright {}_ \chi \mathcal{A}_ \phi \curvearrowleft M_\phi(k). $$ Here, we had $$ {}_ \chi \mathcal{A}_ \phi = C_c \biggl( K \backslash G(\mathbb{A}_ K) / \prod_{v \neq 0, \infty} G(\mathcal{O}_ v) \times (I_0, \chi) \times (I_\infty^{\mathrm{op},+}, \phi) \biggr) \cong k[\Omega]. $$ Given any extension of $\chi$ to $\tilde{\chi} \colon M_\phi(k) \to \mathbb{C}^\times$, we can now impose this equivariance condition and get a one-dimensional space.