Very related is this post by mr_e_man.
Call the alternative conditions $(2')$ and $(4')$.
Condition $(2)$ merely says that $\hat V = \alpha V$ for some scalar $\alpha$, whence
$$
V + \hat V = (1+\alpha)V,\quad V - \hat V = (1-\alpha)V
$$
so if $\alpha\ne\pm1$ then $V$ is both even and odd, and hence 0, which contradicts (1). Thus $\alpha = \pm1$, and $(2)$ is just a silly way of saying that $V$ has homogeneous parity. (3) Then easily follows and is entirely unecessary.
(1) and (4') together make all other conditions unnecessary.
Our revised lists of conditions are
$$\begin{cases}
(1) & V\text{ is invertible}, \\
(2'')& V\text{ is even or odd}, \\
(4) & Vu\tilde V\text{ is a vector when $u$ is}.
\end{cases}
\quad\text{or}\quad
\begin{cases}
(1) & V\text{ is invertible}, \\
(4')& \hat VuV^{-1}\text{ is a vector when $u$ is}.
\end{cases}
$$
It suffices to show (a) that the first set of conditions implies that $V\tilde V$ is a scalar; and (b) that the second set implies that $V$ is a versor.
However, (a) is false per the counterexample given by mr_e_man in the comments. Consider $1 + I \in \mathbb G^{3,1}$, $I = e_1e_2e_3e_4$. Clearly this is even and has inverse $\frac12(1 - I)$. Noting that $(1 + I)^\sim = 1 + I$, for any vector $u$ we have
$$
(1 + I)u(1 + I) = u(1 - I)(1 + I) = u(1 - I^2) = 2u.
$$
But
$$
(1 + I)(1 + I) = 1 + 2I + I^2 = 2I
$$
is not a scalar.
Actually, we can use this idea to generate infinitely many counterexamples in many different signatures. All we need is $p+q = 2k$, $I^2 = -1$, and $\tilde I = I$. The last two conditions together imply that $q$ is odd; together with the first this implies that $p$ is odd; and considering the third again we get that $p + q$ must be a multiple of 4.
Thus, all the following and more yield counterexamples:
$$
(3,1)\quad(1,3)\quad(1,7)\quad(7,1)\quad(3,5)\quad(5,3)
$$$$
(9,3)\quad(3,9)\quad(7,5)\quad(5,7)\quad(11,1)\quad(1,11)
$$
In contrast, claim (b) is a standard result:
The group
$$
\Gamma = \{V \;:\; V^{-1}\text{ exists and }\hat VuV^{-1}\text{ is a vector for all vectors }u\}
$$
is called the Lipschitz group. For any vector $u$ note that
$$
(\hat VuV^{-1})^2 = \hat VuV^{-1}\hat VuV^{-1} = (V\hat u\hat V^{-1})^\wedge\hat VuV^{-1} = Vu(\hat V)^{-1}\hat VuV^{-1} = u^2
$$
so that $u \mapsto \hat VuV^{-1}$ is an isometry; thus we have a group homomorphism $\psi : \Gamma \to \mathrm O(p,q)$ given by $\psi_V(u) = \hat VuV^{-1}$. Now note that
$$
\hat VuV^{-1} = u \iff \hat V = uVu^{-1}.
$$
If this condition holds for all $u$ then we see that $V$ is a scalar (by, say, expanding $V$ in an orthogonal basis and choosing $u$ wisely). In other words, $\ker(\psi) = \mathbb R^\times$. Finally, we know that if $V$ is a vector then $\psi_V$ is a reflection; by the Cartan–Dieudonné theorem and the kernel result above we have that every element of $\Gamma$ is (proportional to) a product of at most $p+q$ vectors.
