Given an ordinal with a particular cofinality, can we find other cofinal subsets with order type and cofinality at least that large? Followup to: If $\omega^\alpha$ has cofinality $\omega^\beta$, and $\beta\le\gamma\le\alpha$, does it also have a cofinal subset of order type $\omega^\gamma$?
Say we have an ordinal $\omega^\alpha$ and suppose it has cofinality $\omega^\beta$, i.e., $\omega^\beta$ is the smallest order type of a cofinal subset.  (Note that $\omega^\beta$ must be a regular ordinal and in particular an initial ordinal.)  Pick $\gamma$ such that $\beta\le\gamma\le\alpha$, and such that, in addition, $\mathrm{cf}(\omega^\gamma)\ge \omega^\beta$ as well.
The question is: Must $\omega^\alpha$ have a cofinal subset of order type exactly $\omega^\gamma$?  If false as stated, is it at least true if we restrict to $\mathrm{cf}(\omega^\gamma)=\omega^\beta$?  What if we restrict to $\beta=1$?
Edit: As Andres Caicedo points out in the comments, we have to restrict to $\mathrm{cf}(\omega^\gamma)=\omega^\beta$.  But I don't currently see how to prove it true under that condition.
Notes: I originally asked this without the restriction on the cofinality of $\omega^\gamma$; Asaf Karagila pointed out that the statement was obviously false for cofinality reasons since cofinality is far from monotonic.  So, I've added a cofinality restriction to avoid that issue. :)
Thank you!
 A: OK, I think I see how to do this now.  It's already been explained in the comments why this is impossible unless in fact $\mathrm{cf}(\omega^\gamma)=\omega^\beta$, so that answers that question.
So say $\mathrm{cf}(\omega^\gamma)=\mathrm{cf}(\omega^\alpha)=\omega^\beta$, $\gamma\le\alpha$.  Let $B\subseteq\omega^\gamma$ be cofinal of type $\omega^\beta$, $B'\subseteq\omega^\alpha$ cofinal of type $\omega^\beta$, and let $\phi:B\to B'$ be the isomorphism.
Induct on the elements of $B$ to build $\psi:B\to B'$ and $f:\omega^\alpha\to\omega^\gamma$; assume $\psi(c)$ is defined for $c<b$ and $f(x)$ is defined for $x<\psi(c)$ for some $c<b$.  Let $x_0$ be the supremum of all $c<b$ within $\omega^\gamma$, so it's the smallest element which hasn't had $f$ defined yet, and let $x'_0$ be the supremum of all $f(x)$ defined so far.
Then for our current $b\in B$, pick some $\psi(b)\in B'$ such that $\psi(b)\ge x'_0+b$ and $\psi(b)>\psi(c)$ for all $c\in B$ with $c<b$.  Then in particular, $\psi(b)\ge x'_0+(-x_0+b)$, so it's possible to choose $f$ so as to map the interval $[x_0,b)$ into the interval $[x'_0,\psi(b))$.
When all this is done, we have an embedding $f:\omega^\gamma\to\omega^\alpha$; and since we must have $\psi(b)\ge \phi(b)$ for $b\in B$, it follows that the image of $f$ is cofinal.
