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 $n \geqslant 0$. Write $W(t)$ rather than $W_t$.

Suppose that $f$ with the desired property exists, that both $W(t)$ and $V(t)$ take every real value, and that $V(t)$ is not monotone on any interval (which, of course, happens with probability one). The argument is divided into a number of steps.

1. For some one-to-one function $\phi$, we have
$$ V(f(t)) = \phi(W(t)) . $$
Indeed: for every $x$ find an arbitrary $T(x)$ such that $W(T(x)) = x$, and set $\phi(x) = V(f(T(x))$. Then $V(f(t)) = \phi(x)$ if and only if $V(f(t)) = V(f(T(x)))$, that is, $W(t) = W(T(x))$, that is, $W(t) = x$.

2. The function $\phi$ is in fact strictly monotone. Indeed: for a given $x \in \mathbb{R}$, there is $T(x) \geqslant 0$ such that $W(T(x)) = x$ and for every neighbourhood $I$ of $T(x)$, the set $W(I)$ contains a right neighbourhood of $x$. Thus, $\limsup_{y \to x} \phi(y) \leqslant \limsup_{t \to T(x)} \phi(W(t)) = \limsup_{t \to T(x)} V(f(t)) = V(f(T(x))) = \phi(W(T(x))) = x$. A similar argument shows that $\liminf_{y \to x} \phi(y) \geqslant \phi(x)$. Thus, $\phi$ is continuous at $x$. Since $x$ is arbitrary, $\phi$ is continuous, and hence (being one-to-one) strictly monotone.

3. With no loss of generality we assume $\phi$ is strictly increasing. The other case is dealt with in a similar manner.

4. 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 $$T_{2n+2} = \inf\{t > T_{2n+1} : W(t) > M(T_{2n+1})\}$$
In other words $T_{2n+1}$ is the first time $W(t)$ exceeds its current infimum after $T_{2n}$, and $T_{2n+2}$ is the first time $W(t)$ exceeds its current supremum after $T_{2n+1}$. Then $T_n$, $W(T_{2n})$ and $-W(T_{2n+1})$ all go to infinity as $n \to \infty$. (It is good to make a picture here.)

5. The tail-equivalence class of $(T_n)$ does not depend on $T_0$. Indeed: suppose that $T_0' > T_0$ and $T_n'$ is defined in a similar way as $T_n$, but with $T_0$ replaced by $T_0'$. Clearly, $T_0' \in [T_{2n}, T_{2n+2})$ for some $n$. If $T_0' < T_{2n+1}$, then $T_1' = T_{2n+1}$ and consequently $T_k' = T_{2n+k}$ for $k > 0$. If $T_0' \geqslant T_{2n+1}$, then either $T_1' \in [T_{2n+1}, T_{2n+2})$ and consequently $T_k' = T_{2n+k}$ for $k > 1$, or $T_1' \in [T_{2n+3}, T_{2n+4})$, and consequently $T_k' = T_{2n+2+k}$ for $k > 1$. (Looking at a picture helps a lot here.) Either way, $T_n$ and $T_n'$ are tail-equivalent.

6. Recall that $W(t)$ exceeds its past supremum at $T_{2n}$; that is, there is a sequence $\epsilon_k > 0$ convergent to zero, such that $X(T_{2n} + \epsilon_k) > M(T_{2n})$ for every $k$. Therefore, $f(T_{2n} + \epsilon_k) \notin f([0, T_{2n}])$. This means that $f(T_{2n})$ is one of the endpoints of $f([0, T_{2n})$. A similar argument shows that $f(T_{2n+1})$ is one of the endpoints of $f([0, T_{2n+1}])$.

7. For the next few items, suppose that $\phi$ is unbounded both from below and from above. Choose $n$ large enough, so that
$$\phi(W(T_{2n})) > \sup_{s \in [0, f(0)]} V(s), \qquad \phi(W(T_{2n+1})) < \inf_{s \in [0, f(0)]} V(s).$$
By item 6, $f(T_{2n})$ is one of the endpoints of $f([0, T_{2n}])$, and by the above condition, $f(T_{2n})$ does not belong to $[0, f(0)]$. Therefore, $f(T_{2n})$ is the *right* endpoint of $f([0, T_{2n}])$. Similarly, $f(T_{2n+1})$ is the *right* endpoint of $f([0, T_{2n}])$. This means that $f(T_n)$ is eventually non-decreasing, and for $n$ large enough, $f(T_{2n})$ is the first time $V(s)$ exceeds its past supremum after $f(T_{2n-1})$, and $f(T_{2n+1})$ is the first time $V(s)$ exceeds its past infimum after $f(T_{2n})$.

8. Define the sequence $S_n$ in a similar way as $T_n$, but using $V(t)$ rather than $W(t)$. The previous item shows that $f(T_n)$ and $S_n$ are tail-equivalent.

9. 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. (Again, have a look at the picture.) 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 $W(t)$.

10. By continuity of $f$, $A_n = 1$ implies $C_n = 1$ (but not necessarily *vice versa*). Therefore, $C_n \geqslant A_n$. 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.

11. The probability that, given 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, and additionally the corresponding $\phi$ is unbounded both from below and from above, is necessarily zero.

12. We now turn our attention to the case when $\phi$ is bounded from below or bounded from above. Clearly, it is sufficient to consider the case when $\phi$ is bounded from below. The argument is here more sketchy, but I will try to fill in the details later. 

13. If $\phi(x) \geqslant c$ for every $x$, then $f$ is bounded (for $f$ necessarily takes values in a connected component of $\{s : V(s) \geqslant c\}$). If we define $T_n$ as in item 7, then one can show that the sequences $f(T_{2n})$ and $f(T_{2n+1})$ are eventually monotone, and one of them is eventually increasing, and the other eventually decreasing. (Otherwise, the path of $V_s$ would have an infinite number of oscillations of a fixed size over a finite time horizon, a contradiction with continuity.) With no loss of generality we consider the case where $f(T_{2n})$ is increasing for $n \ge N$, and $f(T_{2n+1})$ is decreasing for $n \ge N$.

14. Since $V(s)$ is not monotone on any interval, it attains a local extremum somewhere in the interior of $f([T_{2N+1}, T_{2N+2}])$. Let $V(s_0) = y_0$ be this local extremum, and let $y_0 = \phi(x_0)$. What we have found above implies that $W(t)$ attains a local extremum equal to $x_0$ in *every* interval $[T_{2n+1}, T_{2n+2}]$, $n = N, N+1, \ldots$ However, with probability one, the local extrema of $W(t)$ are all distinct. Thus, the probability that a function $f$ with the desired property exists, and additionally the corresponding $\phi$ is bounded both from below or from above, is necessarily zero.

The desired result follows.