**Edit:** I remove the trivial part of the first version, according to comment of Alex Degtyarev


Let $M$ be the manifold of all matrices in $M\_{n}(\mathbb{R})$ with fixed rank $0<k<n$. There is a canonical $n-k$ bundle $E$ over $M$. $E$ is the kernel bundle:$$\{(A,v)\in M\_{n}(\mathbb{R})\times \mathbb{R}^{n}\mid Av=0\}$$


We consider the $Z\_{2}$ action on base space $M$ via transpose operation



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> Question: Can one say that: This action can not be lifted to a $\mathbb{Z}\_{2}$- equivariant structure for $E\to M$?
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$\mathbb{Z}_{2}$ -equivariant vector bundles over manifold of rank-$k$ matrices
The transposition action does not lift, at least, "typically". The easiest obstruction is in the case $k=1$; in general, I think it's the same, with a bit more characteristic classes (or, "projective space" replaced with other magic words like "Grassmannian"). So, restrict the bundle to matrices with a fixed image $V=\Bbb R\subset\Bbb R^n$. The bundle in question is essentially the cotautological bundle over the "almost" projective space $\Bbb R^n\smallsetminus0=\operatorname{Hom}(\Bbb R^n,\Bbb R)\smallsetminus0$. But the action takes this subset to matrices whose **kernel** is the fixed space $V^\perp$, to which the restriction of the bundle is trivial.


**Added:**
In fact, the action obviously lifts if $k=0$ or $n$, and does not lift otherwise. For proof, as above, one can stick to the projective space: just restrict the bundle to the set of block-diagonals of the identity $\operatorname{id}\colon\Bbb R^{k-1}\to\Bbb R^{k-1}$ and rank $1$ matrices $\Bbb R^{n-k+1}\to V\subset\Bbb R^{n-k+1}$ with fixed image $V$.


