If $M$ is a standard class model of ZFC isomorphic to $V$, then is $M = V$? Consider the following statement: (T) "If $M$ is a standard class model of ZFC isomorphic to $V$, then $M = V$." The statement (T) is equivalent to: "If the transitive collapse of a standard class model $M$ of ZFC is equal to $V$, then $M = V$."  This is because the transitive collapse of a class $M$ is the unique transitive class that is elementhood-wise isomorphic to $M$.
Here, by standard class model of ZFC I mean a class model of ZFC whose elementhood relation is the real elementhood relation.
Assume that ZFC is consistent.  Does ZFC prove (T)?  Does ZFC disprove (T)?  If no to both, does ZFC with some additional large cardinal axiom disprove (T)?

No.  Define $F:V\to V$ by $\in$-recursion as $F(x)=\{F(y):y\in x\}\cup\{\emptyset\}$.  Clearly $F(x)$ is nonempty for all $x$.  Also, $F$ is injective: if $F(x)=F(x')$, then by induction on $\max(\operatorname{rank}(x),\operatorname{rank}(x'))$ we may assume $F$ is injective on $x\cup x'$.  Since $F(x)=F(x')$ we must have $\{F(y):y\in x\}=\{F(y):y\in x'\}$, but since $F$ is injective on $x\cup x'$ this implies $x$ and $x'$ have the same elements and thus $x=x'$.  Also clearly $y\in x$ implies $F(y)\in F(x)$, and the converse follows from injectivity of $F$.
Taken all together, this shows that $F$ is an isomorphism from $(V,\in)$ to $(M,\in)$ where $M$ is the image of $F$.  But $M\neq V$, since $\emptyset\not\in M$.
