Let $\mathrm{Z}$ be the formula in question, which can be rewritten as:
$(\Diamond\alpha\land\Box(\alpha\rightarrow\Box\alpha))\rightarrow\Box\alpha$.
If I understand correctly "No world sees two distinct worlds, each of which sees no world other than itself"
from the previous answer, then this is a necessary,
but not sufficient, condition for a frame to validate $\mathrm{Z}$.
For example, the frame $F=(\{A,B,C,D\},\{(A,B),(B,C),(A,D)\})$ satisfies the condition,
yet $\mathrm{Z}$ is invalid on $F$ for the valuation $V(\alpha)=\{A,B,C\}$.
A necessary and sufficient condition for a frame to validate $\mathrm{Z}$
is that the following holds for every world $w$:
$C(\mathrm{Z})$: If $w$ sees $w'$ then every $w''\not=w'$ seen by $w$ can see $w'$
in a finite number of steps through worlds seen by $w$.
Proof: Let $S(w)$ be the set of worlds seen by $w$ and $w_0\in S(w)$.
For any $w_i\in S(w)$ for which $w_i\not=w_0$,
let $L_{w_0}(w_i)>0$ be the length of the shortest path from $w_i$ to $w_0$ through worlds from $S(w)$;
we write $L_{w_0}(w_i)=\infty$ if no such path exists.
Sufficiency: Assume $C(\mathrm{Z})$ holds.
We attempt to invalidate $\mathrm{Z}$ at $w$ and show that this is impossible.
If $S(w)=\varnothing$ then $\mathrm{Z}$ is obviously valid at $w$.
If $w$ sees some world, then to invalidate $\mathrm{Z}$ at $w$ we need $V(\Diamond\alpha,w)=1$,
$V(\Box(\alpha\rightarrow\Box\alpha),w)=1$ and $V(\Box \alpha,w)=0$,
so we need $V(\alpha,w_0)=0$ for some $w_0$ in $S(w)$.
For the remaining $w_i\in S(w)$ (if any), $C(\mathrm{Z})$ implies $0<L_{w_0}(w_i)<\infty$. Then:
For any $w_1\in S(w)$ with $L_{w_0}(w_1)=1$, if $V(\alpha,w_1)=1$ then $V(\alpha\rightarrow\Box\alpha,w_1)=0$
(because $w_1$ sees $w_0$ where $V(\alpha,w_0)=0$), hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$
and $\mathrm{Z}$ is valid at $w$.
So we need to take $V(\alpha,w_1)= 0$ at all such $w_1$.
For any $w_2\in S(w)$ with $L_{w_0}(w_2)=2$, if $V(\alpha,w_2)=1$ then
$V(\alpha\rightarrow\Box\alpha,w_2)=0$
(because $w_2$ sees some $w_1\in S(w)$ with $L_{w_0}(w_1)=1$ where we took $V(\alpha,w_1)=0)$,
hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$ and $\mathrm{Z}$ is valid at $w$.
So we need to take $V(\alpha,w_2)=0$ at all such $w_2$.
We use induction to conclude that to invalidate $\mathrm{Z}$ at $w$ we need to take
$V(\alpha,w_i)=0$ at all $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$,
otherwise $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$ and $\mathrm{Z}$ is valid at $w$.
But since there are no other worlds in $S(w)$ we have $V(\Diamond\alpha,w)=0$,
therefore $\mathrm{Z}$ is still valid at $w$.
Necessity: Assume $w$ sees some $w_0$ and there are also worlds $w_x\in S(w)$ with $L_{w_0}(w_x)=\infty$.
Then we take:
$V(\alpha,w_i)=0$ for all $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$,
$V(\alpha,w_x)=1$ for all $w_x\in S(w)$ with $L_{w_0}(w_x)=\infty$,
as well as $V(\alpha,w_y)=1$ for any $w_y$ seen by such $w_x$.
This is possible because no such $w_x$ sees a $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$
(otherwise it would itself have $L_{w_0}(w_x)<\infty$).
For this valuation $V(\alpha\rightarrow\Box\alpha,w_x)=1$,
hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=1$ continues to hold, as well as $V(\Box\alpha,w)=0$.
But this time $V(\Diamond\alpha,w)=1$ and $\mathrm{Z}$ is invalid at $w$.