Why are "innermorphisms" not useful? A commonly studied type of linear function in geometric algebra (and more generally, exterior algebra) is the outermorphism.  For reference, here's Wikipedia's definition:

Let $f$ be an $\Bbb R$-linear map from $V$ to $W$. The outermorphism of $f$ is the unique map $\underline{\mathsf{f}} : \Lambda(V) \to \Lambda(W)$ satisfying
$$ \underline{\mathsf{f}}(x) = f(x)\\ 
 \underline{\mathsf{f}}(A \wedge B) = \underline{\mathsf{f}}(A) \wedge \underline{\mathsf{f}}(B)\\
 \underline{\mathsf{f}}(A + B) = \underline{\mathsf{f}}(A) + \underline{\mathsf{f}}(B)\\
 \underline{\mathsf{f}}(1) = 1$$
for all vectors $x$ and all multivectors $A$ and $B$, where $\Lambda(V)$ denotes the exterior algebra over $V$.

Why do we not also study linear mappings which preserve the inner product: "innermorphisms"?  We could define it analogously.  The only change in the definition above would be the second equation would read $$\underline{\mathsf{f}}(A \cdot B) = \underline{\mathsf{f}}(A) \cdot \underline{\mathsf{f}}(B)$$
There are at least $2$ examples of these functions: both reflections and rotations preserve the inner product (and have all of the other properties listed above).
Is it that these are the only ones?  Or that the only functions which are innermorphisms are also the orthogonal functions and thus we are already studying them?
I'm just not sure why outermorphisms are so useful (and they are), but that analogous functions which preserve the inner product are apparently not.
 A: We do, but for historical reasons they are called (linear) isometries.
A: Setup
I suppose $V$ and $W$ are finite dimensional real vector spaces for convenience. I will use $F$ in place of $\underline{\mathsf{f}}$. To extend the inner product of vectors to all multivectors, I will follow the convention of Geometric Algebra book by MSE's own Alan Macdonald, where $\cdot$ is used to denote the "left contraction" ( sometimes denoted with the symbol $\rfloor$ ).

Outermorphisms
Why defined?
As pointed out by mr_e_man in a comment, this yields a very natural extension of a linear map $f:V\to W$ that is defined on all multivectors in $\Lambda(V)$. It's natural because it preserves the induced exterior algebra, as mentioned on Wikipedia.
Completely linear?
In a different comment, mr_e_man raised the question of whether an outermorphism must be linear from $\Lambda(V)\to\Lambda(W)$. Since $F$ is additive. This makes it linear over $\mathbb Q$, but not $\mathbb R$ (see, for example, this related MSE question).
I'll have to think more about this, but it may be forced that $F(a)=a+g(a)I$ where $g(a)$ is a scalar function (which is $0$ on every rational) and $I$ is a pseudoscalar for the image of the original linear map $f$.
To avoid the weird cases, it suffices to explicitly assume that $F$ is linear on scalars, since wedge products with scalars covers everything else. (Alternatively, throw away the axiom of choice and work in a model of $\mathsf{ZF}$ where $\mathbb R$-linearity is forced.)
An alternative definition
The standard definition in the question post begins with a linear map $f:V\to W$ and then uses some properties to extend it to all of $\Lambda(V)$ (with a new codomain of $\Lambda(W)$).
However, we could derive $f$ from $F$, suitably defined. We could say that an outermorphism is a map $F:\Lambda(V)\to\Lambda(W)$ satisfying
$$ F(A \wedge B) = F(A) \wedge F(B)\\
 F(A + B) = F(A) + F(B)\\
 F(a) = a\text{ for scalars }a\\
F(\mathbf v)\in W\text{ for }\mathbf v\in V$$
(Actually, I'm pretty sure "$F(a)=aF(1)$ and $F(1)\ne0$" could be used to replace "$F(a) = a$", but that requires some algebra.)

"Innermorphisms"
Another meaning
As an aside, "innermorphism" is sometimes used to mean "inner automorphism" (as in this MSE answer or these review notes) or something related (as in the GAP documentation). Here I am following the lead of the question and examining a very different meaning by analogy with the outermorphisms above.
Definition
In analogy to the definition of outermorphism above, how about

An innermorphism is a map $F:\Lambda(V)\to\Lambda(W)$ satisfying $$
 F(A \cdot B) = F(A) \cdot F(B)\\  F(A + B) = F(A) + F(B)\\  F(a) =
 a\text{ for scalars }a\\ F(\mathbf v)\in W\text{ for }\mathbf v\in V$$
(Again, I suspect "$F(a)=aF(1)$ and $F(1)\ne0$" could be used to replace "$F(a) = a$", but I've only verified this for $W$ up to $2$D.)

The core linear map
If we restrict $F$ to $V$, we have a linear map $f$. But, unlike in the outermorphism case, it's not arbitrary since $F(\mathbf v)\cdot F(\mathbf w)=F(\mathbf v\cdot \mathbf w)=\mathbf v\cdot \mathbf w$. Therefore, $F$ is an orthogonal transformation on $V$ (see, e.g. these notes).
Extending to multivectors
In the general case, it seems that (unlike outermorphisms) innermorphisms are not defined by their core linear map on ($1$-)vectors.
For example, suppose $W$ is at least four dimensional, with orthonormal set $\mathbf w_1,\ldots,\mathbf w_4$. And suppose $V$ is at least two dimensional, with $\mathbf e_1,\mathbf e_2$ an orthonormal pair so that $F(\mathbf e_i)=\mathbf w_i$ for $i=1,2$.
Then $F(\mathbf e_2)=F(\mathbf e_1\cdot(\mathbf e_{1}\wedge\mathbf e_2))=F(\mathbf e_1)\cdot F(\mathbf e_{1}\wedge\mathbf e_2)=\mathbf w_1\cdot F(\mathbf e_{1}\wedge\mathbf e_2)$ and $F(\mathbf e_1)=F(-\mathbf e_2\cdot(\mathbf e_{1}\wedge\mathbf e_2))=-F(\mathbf e_2)\cdot F(\mathbf e_{1}\wedge\mathbf e_2)=-\mathbf w_2\cdot F(\mathbf e_{1}\wedge\mathbf e_2)$.
This does not allow $F(\mathbf e_{1}\wedge\mathbf e_2)$ to be, say, $\mathbf w_1\wedge\mathbf w_2+\mathbf w_1\wedge\mathbf w_3$, but seems to allow it to be things like $7+\mathbf w_1\wedge\mathbf w_2+5(\mathbf w_3\wedge\mathbf w_4)$. And if $W$ has more dimensions, then a term like $\mathbf w_3\wedge\mathbf w_4\wedge \mathbf w_5$ is possible, too.
Why not defined?
In summary, innermorphisms restrict the core linear map to be orthogonal, and do not do enough to define the map on arbitrary multivectors (which is the point of outermorphisms), so they probably don't need to be defined.
