Explicit invariants (under change of basis) of maps $V \to V \otimes V$. It is standard to construct numbers associated to a linear transformation $f: V \to V$  of a finite-dimensional vector space which are invariant under change of basis.  The coefficients of the characteristic polynomial are such, and it is quite simple to see for example that the trace is invariant by equating it with the map which takes $\sum v_i \otimes w_i \in V \otimes V^* \cong Hom(V, V)$ and then evaluates $\sum w_i(v_i)$.
My question is: can one explicitly associate quantities to maps $V \to V \otimes V$ which are invariant under change of basis?  A simple dimension count indicates there should be plenty of invariants, since the orbits will have dimension roughly $d^3 - d^2$, where $d$ is the dimension of $V$, but I have yet to construct a single one.
 A: The answer depends on what you mean by "quantities"! As José demonstrated, there are no polynomial absolute invariants, essentially because of homogeneity, however, there are plenty of rational ones, obtained as fractions $F/G,$ where $F$ and $G$ are polynomial relative invariants of the same weight $k:$
$$ F(g\cdot f)=(\det g)^k F(f),\  G(g\cdot f)=(\det g)^k G(f), \quad g\in GL(V).$$
Your dimension count is one of the common false believes. In fact, sanity is restored when working with rational functions, for
$$ \operatorname{tr\ deg} K(X)^G=\operatorname{tr\ deg} K(X)-\dim O_x,$$
where $X$ is an irreducible algebraic variety over an alg. closed field $K$ of char 0 with an action of an algebraic group $G$, $O_x=G\cdot x$ is a generic orbit, $G_x$ is the stabilizer of $x$, $\dim O_x=\dim G-\dim G_x.$ In fact, rational invariants always separate generic orbits.1  What goes wrong with polynomial invariants is that orbits need not be Zariski closed. In your example, due to the presence of dilations, Zariski closure of every orbit contains zero. A $G$-invariant polynomial function $F$ is constant on any orbit $O$ and hence its value at any point $x\in O$ is equal to $F(0),$ so $F$ is constant. 
Another way to resolve the issue is to replace the group $G$ with its subgroup $[G,G]$, which is the inter-section of the kernels of all 1-dimensional representations of $G$, thus replace $Gl(V)$ with $SL(V)$ om the example. After the $GL(V)$-equivariant identification $Hom(V,V\otimes V)\simeq V^*\otimes V\otimes V$ and polarization, which replaces a homogeneous degree $d$ polynomial function on $W$ with a multilinear map with $d$ arguments $W^\otimes d\to K$, the question reduces to finding multilinear $SL(V)$-invariants. Classical invariant theory shows that they are all obtained by composing $GL(V)$-invariant contractions $V^* \otimes V\to K,$ $\xi\otimes v\mapsto \xi(v)$ and expansions $K\to V^*\otimes V,$ $1\mapsto \sum e_i^*\otimes e_i$, permutations, symmetrizations, antisymmetrizations, and the $SL(V)$-invariant determinants $V^{\otimes k} \to K,$ $v_1\otimes\ldots\otimes v_k\mapsto \det[v_1|\ldots|v_k],$ $k=\dim V.$ Explicit formulas are a bit messy.

Footnotes 
1See Vinberg and Popov's article on invariant theory in Algebraic Geometry 4 volume of the yellow Russian Math Encyclopaedia.   
A: No nontrivial linear one exists, I'm afraid.
The underlying reason why the trace exists is that there is a $\mathrm{GL}(V)$-equivariant endomorphism of $V$: namely, the identity endomorphism.
By contrast, there cannot be any $\mathrm{GL}(V)$-equivariant linear maps $V \to V \otimes V$.
For definiteness, let us consider $V$ a real vector space.
Consider the action of the $\mathbb{R}^\times$ subgroup of $\mathrm{GL}(V)$ consisting of scalar matrices.  Then if $f:V \to V \otimes V$ is linear
$$f(\lambda v) = \lambda f(v)$$
for any $v \in V$ and $\lambda \in \mathbb{R}^\times$.
On the other hand, equivariance would say that
$$ \lambda \cdot f(v) = \lambda^2 f(v).$$
You can only reconcile both if $f$ is identically zero.
Edit
Unwisely, I had assumed linearity.  (The OP did mention other polynomial invariants.)  As pointed in comments to this answer, there are of course rational invariants.
A: This is essentially a rephrazing of Victor Protsak's reply (which I read too late).
In characteristic zero the answer is in principle given by classical invariant theory. Let $(e^i)$, $i=1,...,n$ be a basis for V. Then
coordinates on the space $Hom(V,V \otimes V)$ are given by tensors $c^i_{jk}$. Such a
tensor sends the vector $\lambda_i e^i$ to $c^i_{jk}\lambda_i e^j \otimes e^k$ (using the summation condition for repeated indices).
In addition let $\epsilon^{i_1,...,i_n}$ be the tensor which is $1$ if the $i_1,...,i_n$
form an even permutation of $1,...,n$, which is $-1$ for an odd permutation and which is zero otherwise. Similarly for $\epsilon_{i_1,...,i_n}$. 
SL(V) invariants are now given by expressions in $c,\epsilon$ which have no free indices
(it is possible to do this via graphs). One has to be careful since many expressions
are zero for symmetry reasons.
There are clearly no linear invariants. If $dim(V)=2$ then I believe $\epsilon^{kl}c^i_{kj}c^j_{il}$ is an honest quadratic invariant. Another one
is $\epsilon^{kl}c^i_{ik}c^j_{lj}$. I.e. the determinant of the left partial
trace with the right partial trace.
EDIT:
The two dimensional case can also be viewed in a less abstract way. If $dim V=2$ then
$V\cong V^\ast$ as $SL(V)$ representation. Furthermore by Clebsch Gordan we have $V^* \otimes V\otimes V\cong
S^3V\oplus V\oplus V$. This is the problem of finding the generating concomitants for binary cubic forms. The answer can be found in Grace and Young. In this setting I
see only one quadratic invariant, namely the determinant between the two copies of V.
This means that the two quadratic invariants identified above are linearly
dependent which does not appear obvious. But they are indeed. If you write them
out explicitly you get (up to sign)
$c^1_{11}c^1_{12}-c^1_{21}c^1_{11}+c^1_{12}c^2_{12}-c^2_{21}c^1_{21}+c^2_{12}c^2_{22}
-c^2_{22}c^2_{21}$
A: Jose's argument proves that there are no polynomial invariants, i.e. no invariants that are polynomial in the matrix coefficients with respect to a given basis.  Victor suggests that there should be rational invariants.  Here's an easy one.  Let $\alpha : V \to V \otimes V$ be your function.  Then there are roughly four (not necessarily symmetric) bivectors $\beta: k \to V\otimes V$ by tracing $\alpha^{\otimes 2} : V^{\otimes 2} \to V^{\otimes 4}$ in various different ways.  For generic $\alpha$, this bivector $\beta$ will be nondegenerate, i.e. it will have an "inverse" $\beta^{-1}: V\otimes V \to k$, defined by declaring that one of the traces of $\beta\otimes \gamma$ is the identity map $V \to V$.  If $\beta^{-1}$ exists, then it is a (nonsymmetric) inner product on $V$.  Now, $\alpha$ determines two natural vectors $\operatorname{tr}_L(\alpha),\operatorname{tr}_R(\alpha)\in V$, by tracing in the two different ways.  So some invariants are the squared lengths of $\operatorname{tr}_L(\alpha),\operatorname{tr}_R(\alpha)$ and their two inner products with respect to $\beta^{-1}$.
A: I've been thinking about the case when $V$ is $2$-dimensional. Here, I claim, is a parametrization of the $GL(V)$-orbits of 'generic' linear maps $F:V\to V\otimes V$:
To numbers $p, a, b, c$ we associate a map which acts on a basis $v, w$ by:
$F(v)= -p(v\otimes v)+ w\otimes w$
$F(w)= (v\otimes w-w\otimes v)+ a(v\otimes v)+b(v\otimes w+w\otimes v)+ c(w\otimes w)$.
To put a generic $F$ in this form, write it as $F_++F_-$, symmetric plus antisymmetric. $F_-$ is given by $x\mapsto v\otimes x-x\otimes v$ for a unique $v\in V$. Assume $v\ne 0$. 
Choose $w$ such that $v,w$ is a basis, but remember we are free to replace $w$ by $sv+tw$ for any scalars $s$ and $t\ne 0$. Think of the symmetric tensor $F_+(v)$ as a homogeneous quadratic polynomial in indeterminates $v,w$. Assume it has a $w^2$ term. Replacing $w$ by suitable $tw$ we can make that term $w^2$. Now replacing $w$ by suitable $sv+w$ ("completing the square") we can eliminate the $vw$ term. So $F_+(v)= -pv^2+w^2=-p(v\otimes v)+ w\otimes w$ for some $p$. And $F_+(w)= av^2+2bvw+cw^2=a(v\otimes v)+b(v\otimes w+w\otimes v)+ c(w\otimes w)$ for some $a,b,c$. This gives the formulas above for $F$.
I used up all the choices I had except that $w$ can still be changed to $-w$, which would change $p,a,b,c$ to $p,-a,b,-c$. So the "parametrization" is actually two to one.
