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This answer is incorrect, since it doesn't work correctly at successors of limits. But this is possible to fix. I'll edit and repost.

Yes. Just take any countable $\omega$-nonstandard model $M\models\text{ZFC}$. In this case, once $\alpha$ is above the standard finite numbers, we can biject $V_{\alpha+2}^M-V_{\alpha+1}^M$ with $V_{\alpha+1}^M-V_\alpha^M$, since these will both be countably infinite sets. Putting these all together, and fixing the standard hereditarily finite sets, we get a permutation $j$ of $M$ that sends $V_{\alpha+1}^M$ exactly to $V_\alpha^M$ for every infinite ordinal $\alpha$ (as well as for all the nonstandard finite numbers $\alpha$).

One can also do this with uncountable $\omega$-nonstandard models, provided that the sizes match up, which is not difficult to arrange. For example, it will work in any $\omega_1$-like $\omega$-nonstandard model, since again all the nonstandard initial segments are countably infinite, and it also works in any saturated model, since all the nonstandard initial segments will again have the same size.

The answer is no.

If $j$ maps each $V_{\omega+n+1}^M$ onto $V_{\omega+n}^M$ for $n\in\omega^M$, then $j[V_{\omega+\omega}^M]=V_{\omega+\omega}^M$ and so it cannot be that $j[V_{\omega+\omega+1}^M]=V_{\omega+\omega}^M$ also, if $j$ is one-to-one, since all the elements in the target are already used.

Yes. Just take any countable $\omega$-nonstandard model $M\models\text{ZFC}$. In this case, once $\alpha$ is above the standard finite numbers, we can biject $V_{\alpha+2}^M-V_{\alpha+1}^M$ with $V_{\alpha+1}^M-V_\alpha^M$, since these will both be countably infinite sets. Putting these all together, and fixing the standard hereditarily finite sets, we get a permutation $j$ of $M$ that sends $V_{\alpha+1}^M$ exactly to $V_\alpha^M$ for every infinite ordinal $\alpha$ (as well as for all the nonstandard finite numbers $\alpha$).

One can also do this with uncountable $\omega$-nonstandard models, provided that the sizes match up, which is not difficult to arrange. For example, it will work in any $\omega_1$-like $\omega$-nonstandard model, since again all the nonstandard initial segments are countably infinite, and it also works in any saturated model, since all the nonstandard initial segments will again have the same size.

This answer is incorrect, since it doesn't work correctly at successors of limits. But this is possible to fix. I'll edit and repost.

Yes. Just take any countable $\omega$-nonstandard model $M\models\text{ZFC}$. In this case, once $\alpha$ is above the standard finite numbers, we can biject $V_{\alpha+2}^M-V_{\alpha+1}^M$ with $V_{\alpha+1}^M-V_\alpha^M$, since these will both be countably infinite sets. Putting these all together, and fixing the standard hereditarily finite sets, we get a permutation $j$ of $M$ that sends $V_{\alpha+1}^M$ exactly to $V_\alpha^M$ for every infinite ordinal $\alpha$ (as well as for all the nonstandard finite numbers $\alpha$).

One can also do this with uncountable $\omega$-nonstandard models, provided that the sizes match up, which is not difficult to arrange. For example, it will work in any $\omega_1$-like $\omega$-nonstandard model, since again all the nonstandard initial segments are countably infinite, and it also works in any saturated model, since all the nonstandard initial segments will again have the same size.

Yes. Just take any countable $\omega$-nonstandard model $M\models\text{ZFC}$. In this case, once $\alpha$ is above the standard finite numbers, we can biject $V_{\alpha+2}^M-V_{\alpha+1}^M$ with $V_{\alpha+1}^M-V_\alpha^M$, since these will both be countably infinite sets. Putting these all together, and fixing the standard hereditarily finite sets, we get a permutation $j$ of $M$ that sends $V_{\alpha+1}^M$ exactly to $V_\alpha^M$ for every infinite ordinal $\alpha$ (as well as for all the nonstandard finite numbers $\alpha$).

Yes. Just take any countable $\omega$-nonstandard model $M\models\text{ZFC}$. In this case, once $\alpha$ is above the standard finite numbers, we can biject $V_{\alpha+2}^M-V_{\alpha+1}^M$ with $V_{\alpha+1}^M-V_\alpha^M$, since these will both be countably infinite sets. Putting these all together, and fixing the standard hereditarily finite sets, we get a permutation $j$ of $M$ that sends $V_{\alpha+1}^M$ exactly to $V_\alpha^M$ for every infinite ordinal $\alpha$ (as well as for all the nonstandard finite numbers $\alpha$).

One can also do this with uncountable $\omega$-nonstandard models, provided that the sizes match up, which is not difficult to arrange. For example, it will work in any $\omega_1$-like $\omega$-nonstandard model, since again all the nonstandard initial segments are countably infinite, and it also works in any saturated model, since all the nonstandard initial segments will again have the same size.