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Write $d=\mathrm{diag}({\varpi }^{-(n-1)},\dots ,1)$ for the diagonal matrix called $\delta$ in this question. If $v\in V^K$ and $k_i=d^{i}k{d}^{-i}\in K_i$, thenThus $\rho (d^i)$ maps the fixed-vector space $V^K$ into $V^{K_i}$. Its inverse $\rho (d^{-i})$ gives the reverse map, so

For an upper-unipotent matrix $u$, inverse conjugation changes each entry above the diagonal byEvery such factor tends to zero as $i\to \infty$. Because $U_0$ is compact, its finitely many coordinate functions are uniformly bounded; because it is open, it contains a neighbourhood of the identity in $U$. Therefore this contraction is uniform on $U_0$ and gives $d^{-i}{U}_0{d}^i\subseteq U_0$ for every sufficiently large $i$. The diagonal group commutes with $d$, soConjugating this inclusion gives $K\subseteq K_i$ for all sufficiently large $i$. This is an eventual inclusion; arbitrary $U_0$ need not give a monotone sequence at every earlier step.

Consequently $V^{K_i}\subseteq V^K$ eventually. Admissibility makes these spaces finite-dimensional, while the isomorphism above makes their dimensions equal. HenceThis proves contraction and admissible fixed spaces by both inclusion and dimension, not by an assumption that conjugate fixed spaces are literally equal from the outset.