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Mar 24 at 11:13 comment added Misha Verbitsky This is why I was talking about finite maps! The idea is to take a finite quotient, and then an embedding. Though the cyclic group cannot act by permutation of the factors, and I was asking about the cyclic group action, so I am not sure we cannot have an embedding.
Mar 23 at 11:50 comment added Nick L @Sasha thanks, yes, I realized this and thus deleted. If anyone is interested the (wrong) example was taking f to be the automorphism of $\mathbb{P}^1 \times \ldots \times \mathbb{P}^1$ given by a cyclic permutation of the factors.
Mar 23 at 10:46 comment added Sasha @NickL: In this example the standard equivariant embedding goes to a projective space of dimension $2^n-1$. But it has many invariant projective subspaces, and it could be that projecting out of one of these you could obtain a smaller embedding. P.S. This is a reply to a comment that was here, but disappeared while I was typing...
Mar 23 at 10:12 answer added Nick L timeline score: 1
Mar 22 at 19:21 history asked Misha Verbitsky CC BY-SA 4.0