Let $W:[0,1]\rightarrow\mathbb R$ be standard Brownian motion with $W(0)=0$.

Let $F_n$ denote the collection of all the $2^n$ many piecewise linear continuous functions $f:[0,1]\rightarrow\mathbb R$ such that $f(0)=0$ and $f$ is linear with slope $\pm \sqrt{n}$ on the intervals $[\frac in,\frac{i+1}n]$ for $0\le i<n$.

Let $\psi_n$ denote a uniformly randomly chosen element of $F_n$, i.e., $\mathbb P(\psi_n=f)=2^{-n}$ for each $f\in F_n$.

Let $\phi_n$ denote a uniformly randomly chosen element of $$ \text{arg min}_{f\in F_n}\left(\sup_{0\le x\le 1}|W(x)-f(x)|\right). $$ In other words, $\phi_n$ is an element of $F_n$ that minimizes the sup-norm distance to $W$. More simply, we can say that $\phi_n$ is a nearest walk to Brownian motion.

Question: Do $\phi_n$ and $\psi_n$ have the same distribution? In other words, is $\mathbb P(\phi_n=f)=2^{-n}$ for all $f\in F_n$? I am mostly interested in the case $n\rightarrow\infty$, but will accept a rigorous answer for $n=2$.

Motivation: Donsker's Theorem says that $\psi_n$ converges weakly to $W$, whereas it is clear that almost surely, $\phi_n$ converges to $W$ uniformly.

EDIT: Here is a follow-up question.