I'm checking a calculation with induced Whittaker functionals and have confused myself a little bit. Let $P_{\ast} = M_{\ast} N_{\ast} \subset P = MN \subset G$ be parabolic subgroups of a $p$-adic reductive group $G$ in their Levi decompositions, containing a Borel subgroup $B = TU$. Let $\chi$ be a generic character of $U$, and $(\pi_{\ast}, V)$ be an irreducible, admissible $\chi$-generic representation of $M_{\ast}$ with Whittaker functional $\lambda$. Let $I_{M_{\ast}}^G (\pi_{\ast})$ be the representation of $G$ obtained by normalized parabolic induction from $P_{\ast}$ to $G$. Then
$$\Lambda_{M_{\ast}}^G: I_{M_{\ast}}^G (\pi_{\ast}) \rightarrow \mathbb C$$
$$f \mapsto \int\limits_{U_{w_{M_{\ast}}^G}} \langle f(w_{M_{\ast}}^{G-1}n'), \lambda \rangle \overline{\chi(n')} \space dn'$$
is a Whittaker functional for $I_{M_{\ast}}^G (\pi_{\ast})$, where $w_{M_{\ast}}^G = w_G w_{M_{\ast}}$ is the product of the long Weyl group elements $w_G$ and $w_{M_{\ast}}$ of $G$ and $M$. Here $U_{w_{M_{\ast}}^G}$ is the product of the root groups for those positive roots made negative by $w_{M_{\ast}}^{G-1}$.
We have similarly the Whittaker functional $\Lambda_{M_{\ast}}^M$ of the representation $I_{M_{\ast}}^M (\pi_{\ast})$ of $M$, and if $\pi$ is a generic representation of $M$, we have in the same way the Whittaker functional $\Lambda_M^G$ of $I_M^G (\pi)$.
I expect that these Whittaker functionals are related in a natural way: if $\pi$ is equal to $I_{M_{\ast}}^M \pi_{\ast}$, then by induction in stages we can naturally identify
$$\Phi: I_M^G \pi \xrightarrow{\cong} I_{M_{\ast}}^G \pi_{\ast}$$
$$\Phi(F)(g) = F(g)(1_M)$$
then we ought to have $\Lambda_M^G = \Lambda_{M_{\ast}}^G$.
The calculation so far:
For $F \in I_M^G(\pi)$, we have
$$\Lambda_M^G(F) = \int\limits_{U_{w_{M}^G}} \langle F(w_M^{G-1}n'), \Lambda_{M_{\ast}}^M \rangle \overline{\chi(n')} \space dn'$$
$$ = \int\limits_{U_{w_{M}^G}} \int\limits_{U_{w_{M_{\ast}}^M}} \langle F(w_M^{G-1}n')(w_{M_{\ast}}^{M-1}n'') , \lambda \rangle \overline{\chi(n'')} \space dn'' \overline{\chi(n')} \space dn'$$
while on the other hand,
$$\Lambda_{M_{\ast}}^G(\Phi(F)) = \int\limits_{U_{w_{M_{\ast}}^G}} \langle \Phi(F)(w_{M_{\ast}}^{G-1}n), \lambda \rangle \space \overline{\chi(n)} dn$$
$$ = \int\limits_{U_{w_{M_{\ast}}^G}} \langle F(w_{M_{\ast}}^{G-1}n)(1_M), \lambda \rangle \space \overline{\chi(n)} dn$$
I'm a little bit unsure of how to reconcile the two elements $F(w_M^{G-1}n')(w_{M_{\ast}}^{M-1}n'')$ and $F(w_{M_{\ast}}^{G-1}n)(1_M)$ in the underlying space of $\pi_{\ast}$.
I see now. Under the isomorphism $\Phi: I_M^G (\pi) \xrightarrow{\cong} I_{M_{\ast}}^G \pi_{\ast}$, each $F \in I_M^G(\pi)$ is a function which sends each $g \in G$ to a function $F(g): M \rightarrow V$ in the space $I_{M_{\ast}}^M\pi_{\ast}$. It follows that if $m' \mapsto F(g)(m')$ is the resulting function of $M$, then
$$F(mg)(m') = \delta_P(m)^{1/2} F(g)(m'm). \tag{$g \in G, m,m' \in M$}$$
It follows that $F(w_M^{G-1}n')(w_{M_{\ast}}^{M-1}n'') = F(w_{M_{\ast}}^{M-1}n''w_M^{G-1}n')(1_M)$ (the modulus character is trivial on the derived group of $M$).
The rest is some technicalities of conjugating root subgroups that probably no one is interested in. I am satisfied with my question, but if someone wants to know the rest of the details I will expand on my answer.