We show here how any interpretation of $\operatorname{Tr} A$ when $A : V \to V$ is an isomorphism can be extended to an interpretation of the trace of an arbitrary endomorphism by showing that $\operatorname{Tr} A$ actually only depends on a special induced sub-vector space of $V$.

To begin, let
$$V^{(0)} = \operatorname{domain}(A) = V,\;\; V^{(i+1)} = A\left(V^{(i)}\right),\; \textrm{ and }d^i = \dim V^{(i)}$$
so that $V^{(1)} = \operatorname{Im} A = A\left(V^{(0)}\right)$, $V^{(i+1)} \subseteq V^{(i)}$, and $d^{i+1} \leq d^i$. Let $N \geq 0$ be the smallest integer s.t. $d^{N+1} = d^N$ and denote this common value by $d$. Let $W := V^{(N)}$.

We prove below that the restriction $A\big\vert_W : W \to W$ of $A$ onto $W := V^{(N)}$ is an isomorphism. Furthermore, $\operatorname{Tr}(A) = \operatorname{Tr}\left(A\big\vert_W\right)$ and it will be clear that $W$ is the unique largest vector subspace $S$ of $V$ on which $A$ restricts to an isomorphism $A\big\vert_S : S \to S$. All of this allows us to conclude that to geometrically interpret $\operatorname{Tr}(A)$, one may restrict their focus to geometrically interpreting the trace of the isomorphism $A\big\vert_W : W \to W$ rather than $A : V \to V$ itself.

This isn't entirely surprising since just as the trace of a matrix does not depend on the "elements off the diagonal", so too does the geometric interpretation of trace not depend on the "space off of $W$." Note that this also gives some geometric intuition about how the trace of a matrix can simultaneously depend only on its diagonal elements while also equaling quantities that non-trivially depend on the whole matrix (such as the sum of its eigenvalues).
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**Proof:** We now prove the above claim. Inductively construct a basis $\left(e_1, \dots, e_{\dim V}\right)$ for $V$ such that for all $i \geq 0$, $\left(e_1, \dots, e_{d^i}\right)$ is a basis for $V^{(i)}$. Let $\left(\varepsilon^1,\dots, \varepsilon^{\dim V}\right)$ be the dual basis of $e_{\bullet}$ and note in particular that:
$$\textrm{(1) whenever }d^{i + 1} < l \leq d^i\textrm{ then }\varepsilon^l\textrm{ vanishes on }V^{(i + 1)}.$$

Since $(e_1, \dots, e_{d^1})$ is a basis for the range of $A$ we may, for any $v \in V^{(0)},$ write
$$A(v) = \varepsilon^1(A(v)) e_1 + \cdots + \varepsilon^{d^1}(A(v)) e_{d^1}$$
so that $A = (\varepsilon^l \circ A) \otimes e_l$ (the sum ranging over $l = 1, \dots, d^1$) and hence
$$\operatorname{Tr}(A) = (\varepsilon^l \circ A)(e_l) = \varepsilon^1(A(e_1)) + \cdots + \varepsilon^{d^1}\left(A\left( e_{d^1} \right)\right)$$
which shows that $\operatorname{Tr}(A)$ actually depends only on the range of $A$ (i.e. $V^{(1)}$).
Now since $e_1, \dots, e_{d^1}$ are (by definition) in $V^{(1)}$, all of $A\left(e_1\right), \dots, A\left(e_{d^1}\right)$ belong to $A\left(V^{(1)}\right) = V^{(2)}$ so that from $(1)$ it follows that
$$\operatorname{Tr}(A) = \varepsilon^1\left(A\left(e_1\right)\right) + \cdots + \varepsilon^{d^2}\left(A\left( e_{d^2} \right)\right)$$

Continuing this inductively $N \leq \dim V$ times shows that
$$\operatorname{Tr}(A) = \varepsilon^1\left(A\left(e_1\right)\right) + \cdots + \varepsilon^{d}\left(A\left(e_d\right)\right)$$
so that $\operatorname{Tr}(A)$ depends *only* on $W = V^{(N)}$.
Since by definition of $N$, the map $A\big\vert_W : W \to W$ is surjective, it is an isomorphism and furthermore, it should be clear that $W$ is the unique largest subspace of $V$ on which $A$ restricts to an isomorphism. $\blacksquare$

As described elsewhere, if you view $A : V \to V$ as a vector field on $V$ in the canonical way then the trace of $A$ is the same as its divergence so in the case where $A$ is an isomorphism there is a pleasing geometric interpretation readily available, which I'll assume that you're comfortable with. However, this interpretation is not satisfactory when $A$ is not surjective, as shown by the example below.

**Example**: If $A : \mathbb{R}^3 \to \mathbb{R}^3$ is such that $\operatorname{Im} A$ is $2$-dimensional and $A$ is volume-increasing (i.e. $\operatorname{div}(A) = \operatorname{Tr}(A) > 0$) then $A$ takes a $3$-d bucket of water (e.g. $\mathbb{R}^3$) and ''compresses it down'' into a $2$-d ''paper'' (e.g. into the 2-d set $\operatorname{Im} A$); but it's not clear (at least to me) how anyone could be expected to say that $A$ has increased the volume of this 3-d space (since $A$ is, after all, "volume-increasing") simply because the quantity $\operatorname{div}(A) = \operatorname{Tr}(A)$ happens to be positive! $\blacksquare$

Nevertheless, the equality $\operatorname{div}(A) = \operatorname{Tr}(A)$ is our best bet at finding a geometric interpretation of trace since it establishes a direct simple relationship between the trace and a readily interpretable quantity.

To summarize, going back to example above, in the case where $A : V \to V$ is an arbitrary linear map we can imagine being given some initial ''bucket of water'' $V = V^{(0)}$ and then imagine $A$ as repeatedly (and eternally) deforming this same water until eventually (i.e. after $N$ iterations) $A$ would have ''pushed'' or ''compressed'' all of $V$ onto some vector subspace $W = V^{(N)}$ (which is also the unique the largest subspace $W$ of $V$ that $A$ maps back onto itself) i.e. all of $V$ would eventually ''flow into'' $W$.
It is at this point that $A$ no longer ''compresses'' this water down by some dimension(s) so that $A$ does nothing more than bijectively move this $d = \dim W$-dimensional water around.
It now makes sense to ask by how much the isomorphism $A\big\vert_W : W \to W$ is increasing or decreasing this $d$-dimensional volume, which is what $\operatorname{Tr}(A) = \operatorname{Tr}\left(A\big\vert_W\right) = \operatorname{div}\left(A\big\vert_W\right)$ represents.

Remark: This may not really answer your question since you stated that "The divergence application of trace is somewhat interesting, but again, not really what we are looking for." Nevertheless, whatever alternative non-divergence based interpretation of the trace of an isomorphism you choose, I hope that this will help you to extend it to the case where the map isn't an isomorphism.