Since "consistent" is a weird notion in the context of second-order set-theories and moreover we can't even directly talk about a second-order theory being true of $V$ within $V$, I think it's usefully demystifying to rephrase the question in a "set-ish" way as follows:

Is it consistent with $\mathsf{ZFC}$ that there is some inaccessible cardinal $\kappa$ such that for every second-order theory $T$ in the language of set theory, if $V_\kappa\models T$ then $V_\alpha\models T$ for some $\alpha<\kappa$?

Note that whether or not $V_\gamma\models S$ for $\gamma\le\kappa$ and $S$ a second-order set theory is detected by the *first-order* diagram of $V_{\kappa+1}$, so the above question does make sense.

As Elliot Glazer observes, there is in this case a simple counting argument we can employ: if there are more than continuum-many inaccessibles, then some pair of inaccessibles $\alpha<\kappa$ have $V_\alpha\equiv_{\mathsf{SOL}}V_\kappa$.

Of course this is somewhat unsatisfying. What we have is a set-theoretic assumption which guarantees **the existence of some appropriate** $\kappa$, but we don't have a concrete property which would **identify** such a $\kappa$. So at this point it's natural to ask:

Is there a natural set-theoretic property - e.g. an already-studied large cardinal property - which guarantees that any $\kappa$ with that property has $V_\alpha\equiv_{\mathsf{SOL}}V_\kappa$ for some $\alpha<\kappa$?

Note that, trivially, such a property would have to be non-second-order-definable. And this pushes us into the realm of very strong properties indeed (see e.g. the discussion here).