This is from Cogdell and Piateski-Shapiro's paper Derivative and L-functions for $GL_n$.
Let $\lambda$ be a non-trivial $\psi$-Whittaker functional. The Whittaker model is defined as$W_v(g)=\lambda(\pi(g)v), v\in V_{\pi},g\in GL_n$. We know that $W_v\in W(\tau,\psi)$ iff there is a compact open subgroup $Y\subset U_n$ such that $$\int_Y W_v(py)\psi^{-1}(y)dy=0,\ \ p\in P_n.$$ I am confused why this can lead to: if writing $p=gu, g\in GL_{n-1}, u\in U_n$, we have $$ \int_YW_v(guy) \psi^{-1}(y)dy=W_v(gu)\int_Y\psi(gyg^{-1})\psi^{-1}(y)dy. $$ I understand it must be using the fact of $\lambda$ being a Whittaker functional, but I am still unable to show $W_v(guy)=W_v(gu)\psi(gyg^{-1}),$ frustrated..
We just need to show $W_v(guy)=W_v(gu)\psi(gyg^{-1})$.
In fact, since $guyu^{-1}g^{-1}\in U_n$, and it is easy to verify $\psi(guyu^{-1}g^{-1})=\psi(gyg^{-1})$, hence $$ W_v(guy)=W_v(guyu^{-1}g^{-1}gu)=\psi(guyu^{-1}g^{-1})W_v(gu)=W_v(gu)\psi(gyg^{-1}). $$