We're working in a Clifford algebra over a non-degenerate $n$-dimensional vector space $V$, and considering various properties a multivector $A$ could have:
(0) $A$ is invertible.
(1) $\tilde AA$ is a scalar (i.e. grade $0$).
(2) $A\tilde A$ is a scalar.
(3) For any vector $v\in V$, $\tilde AvA$ is also a vector (i.e. grade $1$).
(4) For any vector $v\in V$, $Av\tilde A$ is also a vector.
(5) The map $X\mapsto\tilde AXA$ preserves grade. In other words, for any $k$, if $X=\langle X\rangle_k$, then $\tilde AXA=\langle\tilde AXA\rangle_k$.
(6) The map $X\mapsto AX\tilde A$ preserves grade.
(7) $A$ has only even grades, or only odd grades.
(8) $A$ is a product of vectors: $\exists a_1,a_2,\cdots,a_m\in V,\;A=a_1a_2\cdots a_m$.
(9) $A$ is a product of at most $n$ vectors.
I have found some relations among these properties:
(0, 8) imply (9), by the Cartan-Dieudonne theorem. Whether (8) alone implies (9) is this question. In the chat, we have found that (8) implies that $A$ is a product of at most $\tfrac32n$ vectors.
(0, 1, 3, 7) imply (8, 9), since $A$ is then an element of the Lipschitz group.
(5) implies (1, 3), obviously.
(6) implies (2, 4), obviously.
(8) implies (1-7). Suppose $a$ is a vector, and $X$ is a $k$-vector. Then, using the wedge product $a\wedge X=\langle aX\rangle_{k+1}$ and the left contraction $a\,\lrcorner\,X=\langle aX\rangle_{k-1}$, we have
$$aXa=a(X\,\llcorner\,a+X\wedge a)$$ $$=a\,\lrcorner\,(X\,\llcorner\,a)+a\wedge(X\,\llcorner\,a)+a\,\lrcorner\,(X\wedge a)+a\wedge(X\wedge a);$$
the middle two terms are $k$-vectors, and the first term (grade $k-2$) is
$$a\,\lrcorner\,(X\,\llcorner\,a)=a\,\lrcorner\,\big((-1)^{k-1}a\,\lrcorner\,X\big)=(-1)^{k-1}(a\wedge a)\,\lrcorner\,X=0,$$
and the last term (grade $k+2$) is
$$a\wedge(X\wedge a)=a\wedge\big((-1)^ka\wedge X\big)=(-1)^k(a\wedge a)\wedge X=0,$$
so the result $aXa$ is a $k$-vector. It follows by induction that $\tilde AXA=a_m\cdots a_2a_1Xa_1a_2\cdots a_m$ is also a $k$-vector.
(0, 5) imply (6), by general linear algebra: An invertible linear map ($X\mapsto\tilde AXA$) is also invertible when restricted to an invariant subspace (the $k$-vectors), and $\tilde A$ is (proportional to) the inverse of $A$. Note that this works even if $V$ is degenerate.
(5) implies (6). Let $Z=\tilde AXA$. The condition $Z=\langle Z\rangle_k$ is of course equivalent to $\langle Z\rangle_j=0$ for $j\neq k$. Given non-degeneracy, this is also equivalent to $\langle ZY\rangle_0=0$ for all $j$-vectors $Y$ with $j\neq k$. (To see this, just expand everything in terms of an orthogonal basis.) Thus property (5) can be replaced with this:
(5) For all blades $X$ and $Y$ with different grades, $\langle\tilde AXAY\rangle_0=0$.
Then, using the commutativity of the scalar product, we have
$$0=\langle\tilde AXAY\rangle_0=\langle Y\tilde AXA\rangle_0=\langle AY\tilde AX\rangle_0$$
which is just property (6) (since $X$ and $Y$ can be swapped).
(1-6) don't imply (7), as shown by $A=1+e_{123}$ (over a $3$-dimensional space).
The following (un)relations are less important for this question, but....
(1, 2, 7) don't imply (3) or (4), as shown by $A=1+e_{123456}$.
(3, 4, 7) don't imply (1) or (2), as shown by $A=1+e_{1234}$.
(1-4) don't imply (5) or (6), as shown by $A=1+e_1+e_{24}+e_{124}$ and $X=e_{123}$, over $V=\mathbb R^{3,1}$.
(1-4, 7) don't imply (5) or (6), as shown by $A=1+e_{78}+e_{123456}+e_{12345678}$ and $X=e_{17}$, over $V=\mathbb R^{7,1}$.
(2, 7) don't imply (1); see this question and answer.
(2, 3) don't imply (1) or (4), as shown by $A=1+e_1+e_2+e_{12}$, over $V=\mathbb R^{1,1}$.
Question:
(1-7) imply (8)?
We can assume $A$ is even, since (1-8) are unchanged when $A$ is multiplied by any invertible vector.
Note that (1-6) are a set of homogeneous quadratic equations in $A$. And (8) gives parametric equations for $A$, of degree $m\leq\tfrac32n$, in $mn$ variables. So this is a kind of polynomial parametrization problem.
