Call two sequences $(a_n)$ and $(b_n)$ tail-equivalent if there are $p$ and $q$ such that $a_{p+n} = b_{q+n}$ for every $n \geqslant 0$.

Suppose that $f$ with the desired property exists. Then
$$ V_{f(t)} = \phi(W_t) $$
for some one-to-one function $\phi$. It is easy to see that (outside of an event of probability zero) $\phi$ is strictly monotone. With no loss of generality we assume $\phi$ is strictly increasing.

Denote $I_t = \inf_{s \in [0, t]} W_s$ and $M_t = \sup_{s \in [0, t]} W_s$. Let $T_0 > 0$ be chosen arbitrarily (e.g. $T_0 = 1$), and define $$T_{2n+1} = \inf\{t > T_{2n} : W_t < I_{T_{2n}}\}$$ and $$T_{2n+2} = \inf\{t > T_{2n+1} : W_t > M_{T_{2n+1}}\}$$
It is easy to see that the tail-equivalence class of $(T_n)$ does not depend on $T_0$.

Define the sequence $S_n$ in a similar way, using $V_t$ rather than $W_t$.

It is then easy to see that for $n$ large enough, the sequence $f(T_n)$ is increasing and it satisfies a similar recurrence relation, with $W_t$ replaced by $V_t$ (here we use the fact that $\phi$ is strictly increasing). Therefore, $f(T_n)$ and $S_n$ are tail-equivalent.

Define $A_n = 1$ if there are $s, t \in [T_{2n+1}, T_{2n+2}]$ such that $$\text{$s < t$, $W_s > W_{T_{2n}}$ and $W_t < W_{T_{2n+1}}$,}$$
and $A_n = 0$ otherwise. By the strong Markov property, $A_n$ is an i.i.d. sequence of (non-trivial) Bernoulli random variables with some parameter $p \in (0, 1)$.

Define in a similar way $B_n$, using $S_n$ and $V_t$ rather than $T_n$ and $W_t$. Finally, let $C_n$ be a similarly defined sequence for $f(T_n)$ and $V_t$ rather than $T_n$ and $W_t$.

Using continuity of $f$, it is easy to see that $C_n \geqslant A_n$. Furthermore, $B_n$ is tail-equivalent to some $C_n$. At the same time, independence of $W_t$ and $V_t$ implies that $B_n$ and $C_n$ are independent.

The probability that for two independent i.i.d. Bernoulli sequences $A_n$ and $B_n$ (with the same parameter $p$) there is a third sequence $C_n$ such that $C_n \geqslant A_n$ and $C_n$ is tail-equivalent to $B_n$, is easily found to be zero. This proves that the probability that a function $f$ with the desired property exists is necessarily zero.

(There are too many "easy-to-sees" here. Time permits, I will try to expand the relevant parts of the answer.)