TITLE: Dividing by two in the category of vector spaces
QUESTION [23 upvotes]: Does every invertible linear map $M$ between $V \oplus V$ and $W \oplus W$ naturally yield an invertible linear map $L$ between $V$ and $W$?
Here "naturally" means "in an $GL(V) \times GL(W)$-equivariant fashion" [originally I wrote $SL(V)$ here and below, but Todd Trimble and Eric Wofsey pointed out my mistakes]. Cf. the article "Producing New Bijections From Old" by David Feldman and myself, available at http://jamespropp.org/cancel.pdf, in which we raise such questions in the context of maps between finite sets $S$ and $T$ and ask for $Sym(S) \times Sym(T)$-equivariance.
One might try to construct $L$ from $M$ by mapping $V$ to $V \oplus V$ (sending $v$ to $(v,0)$, say), mapping $V \oplus V$ to $W \oplus W$ (by $M$), and then mapping $W \oplus W$ to $W$ (sending $(w,w')$ to $w$, say). But the map $L$ constructed in this fashion need not be invertible; e.g., if $V = W$ and $M$ sends $(v,v')$ to $(v',v)$, then $L$ is the 0-map.
Something cleverer is called for; or maybe by paying attention to $GL(V)$ one can show that no such general construction is possible.
If there is an existing literature on problems of this kind, pointers would be appreciated.
(Note that Conway, Doyle, and Qiu, in http://arxiv.org/abs/math/0605779 and http://arxiv.org/abs/1504.01402, take a slightly different approach to questions of this kind. I'm not sure what a "linear" version of their approach would be.)

REPLY [3 votes]: This comment is about the "natural" part of the question. Suppose we move to Banach spaces so that one naturally expects that if one has a bounded linear isomorphism between $V\oplus V$ and $W\oplus W$, then the corresponding linear isomorphism between $V$ and $W$ should be also bounded. Well, in this case the answer is negative: Gowers and Maurey constructed in "Banach spaces with small spaces of operators (1997)" examples of Banach spaces $V$ which are isomorphic to their "cubes" $V\oplus V \oplus V$ but not to the square  $V\oplus V$. Therefore if we set $W=V\oplus V$, then we have that $V\oplus V$ is isomorphic to $W\oplus W$, while $V$ and $W$ are not isomorphic (as Banach spaces, I mean).