Assume ZFC (and AC in particular) as the background theory.
If $(M,\in^M)$ is a model of ZFC (not necessarily transitive or standard), must there exist a bijection between $M$ and $$\{x \in M \mid (M,\in^M) \models x \mbox{ is an ordinal number}\}?$$
I am also interested in the cases where $M$ is assumed to be a model of ZFC2.
Remark. I think this is different from what was asked here. Please comment if you believe otherwise; I am happy to discuss.
The answer is yes. However, the answer is no if we require the model itself to know the bijection. More specifically, the existence of a class bijection between $V$ and its ordinals is equivalent to the axiom of global choice, and it is consistent that $\mathsf{ZFC}$ holds but global choice fails.
Now, given any (set) model $M$ of $\mathsf{ZFC}$, for each $\alpha$ ordinal of $M$ there is (in $M$) a bijection $f$ between $V_\alpha$ and some ordinal $\beta$ of $M$. Any such $f$ gives us a true bijection between $\hat\beta=\{a\in M\mid M\models a<\beta\}$ and $\hat V_\alpha=\{b\in M\mid M\models b\in V_\alpha\}$, simply by setting $$\hat f=\{(a,b)\in M\times M \mid M \models a<\beta,b\in V_\alpha,\& \,b=f(a)\}.$$ This proves that $$|M|=\left|\bigl(\bigcup_{a\in\mathsf{ORD}^M}V_a\bigr)^M\right|\le|\mathsf{ORD}^M|\sup_{a\in\mathsf{ORD}^M}|\hat V_a|\le |\mathsf{ORD}^M|\sup_{b\in\mathsf{ORD}^M}|\hat b|=|\mathsf{ORD}^M|.$$
The answer is no if we replace $\mathsf{ZFC}$ with $\mathsf{ZF}$. Indeed, we can have transitive models of $\mathsf{ZF}$ that are uncountable and have countably many ordinals, see for instance here.