Say for simplicity's sake I have some finite dimensional vector space $V$ over $\mathbb{R}$.
Question: By varying my choice of inner product, is there a natural manner in which I can picture what is happening to the geometry of my space? How should I think about the 'effect' of endowing $V$ with two different choices of inner product?
I think you've answered your own question in the comments, but we can go into some detail if you want.
If we fix an isomorphism $V \cong \mathbb R^n$, then varying the inner product on $V$ is exactly just changing what vectors are orthogonal to each other (and what their norms are). For example, consider the subspace $\mathbb R^{n-1} \subset \mathbb R^n$ that's given by the first $n-1$ coordinates $(x_1,\ldots,x_{n-1}, 0)$. We can construct an inner product on $\mathbb R^n$ by declaring the first $n-1$ vectors to be orthonormal, and then picking any nonzero vector $v$ in the complement and declaring it to be orthogonal to the others and of norm $1$.
In the standard basis, this looks like we're waving a vector around in the complement and saying that it is what we consider to be of norm $1$ and orthogonal to the other standard basis vectors. It helps to imagine this for $n = 2$ or $n = 3$.
In the general situation "basically the same thing happens". From the point of view of the standard basis, we pick random $n$ vectors that are linearly independent and declare that they are now orthonormal. We can then vary the inner product by varying the vectors we've declared to be orthonormal.
There are some details here that aren't that important, like that two bases can determine the same inner product from this point of view. I think that abstractly we're wandering around in the quotient space $\operatorname{GL}(n) / O(n)$ ($\operatorname{GL}(n)$ because we're picking a basis, and $O(n)$ because that's when two bases determine the same inner product).
Finally note that this is mostly a game of smoke and mirrors. The Gram-Schmidt process means exactly that given any inner product space $(V,g)$, there exists an isometry from it to $(\mathbb R^n, \langle \,\cdot\,,\,\cdot\, \rangle)$, the standard vector space equipped with the standard inner product. Up to isometries, there is only one inner product space of any finite dimension.