From Grothendieck's lemma, we know that all holomorphic vector bundles over the complex projective line are direct sums of line bundles, and so, are $SU(2)$-equivariant. 


I wonder, do there exist non-equivariant vector bundles over complex projective $N$-space, viewed as an $SU(N)$ or $U(N)$ homogeneous space? If so, are any of them also holomorphic vector bundles? What happens for Grassmannians, of flag manifolds, or projective homogeneous spaces? 



Non-equivariant vector bundles over complex projective $N$-space
In some sense, for every $n>1$, most holomorphic bundles on $\mathbb{C}P^n$ are non-equivariant. For instance, let $Z\subset \mathbb{C}P^2$ be a non-empty, zero dimensional closed subscheme that is locally a complete intersection, e.g., a collection of $m>0$ distinct, reduced points. Denote by $\mathcal{I}\_Z$ the (coherent) ideal sheaf of $Z$ in $\mathbb{C}P^2$. Let $d>-3$ be an integer, so that $H^2(\mathbb{C}P^2,\mathcal{O}(d))$ vanishes. Then by "Serre's construction", there exists a short exact sequence of coherent sheaves on $\mathbb{C}P^2$, $$0 \to \mathcal{O}(d) \xrightarrow{q} \mathcal{E} \xrightarrow{p} \mathcal{I}\_Z\to 0,$$
such that $\mathcal{E}$ is locally free of rank $2$. Of course the composition of $p$ with the inclusion, $\mathcal{I}\_Z\subset \mathcal{O}\_{\mathbb{C}P^2}$, is a homomorphism of coherent sheaves, $q':\mathcal{E}\to \mathcal{O}\_{\mathbb{C}P^2}$. The zero locus of $q'$, which depends on $q'$ only up to scaling, is precisely $Z$. 


If $H^0(\mathbb{C}P^2,\mathcal{I}\_Z(-d))$ vanishes, then $q'$ is the unique such homomorphism (up to scaling); of course for $d\geq 0$, always $H^0(\mathbb{C}P^2,\mathcal{I}\_Z(-d))$ vanishes. If $q'$ is unique up to scaling, then the zero locus $Z$ is an invariant of $\mathcal{E}$. This means that $\mathcal{E}$ is not equivariant, since the scheme $Z$ is not invariant under the full group of automorphisms. In fact, once $Z$ consists of $5$ or more general points, the automorphism group of $\mathbb{C}P^2$ mapping $Z$ to itself is trivial, so $\mathcal{E}$ is not equivariant for any nontrivial subgroup of the full automorphism group.


