Answering your more specific question: yes, it$^*$ can be so characterized, in the following way:

• First, start with the axioms of second-order Zermelo set theory with choice (without Replacement). Note that $V_\mu$ is a model of this if $\mu$ is the first limit of inaccessibles - indeed, as long as $\lambda$ is a limit ordinal greater than $\omega$ we have that $V_\lambda$ satisfies second-order Zermelo set theory with choice.

• Next, add "For every ordinal $\alpha$ there is an inaccessible cardinal $>\alpha$."

• Finally, add "There is no ordinal which is a limit of inaccessibles."

This characterizes $V_\mu$ up to isomorphism.

$^*$Or rather, the union of the first $\omega$-many $V_\kappa$s with $\kappa$ inaccessible is so characterizable; in general, if $(\kappa_i)_{i\in\omega}$ is an increasing sequence of inaccessibles there is of course no reason to believe that $\bigcup_{i\in\omega} V_{\kappa_i}=V_{\sup\{\kappa_i:i\in\omega\}}$ is so characterizable.

Answering your more specific question: yes, it$^*$ can be so characterized, in the following way:

• First, start with the axioms of second-order Zermelo set theory with choice (without Replacement). Note that $V_\mu$ is a model of this if $\mu$ is the first limit of inaccessibles - indeed, as long as $\lambda$ is a limit ordinal greater than $\omega$ we have that $V_\lambda$ satisfies second-order Zermelo set theory with choice.

• Next, add "For every ordinal $\alpha$ there is an inaccessible cardinal $>\alpha$."

• EDIT: Now, add "Every set is contained in a set model of ZFC$_2$." Note that since ZFC$_2$ has only finitely many axioms, this is in fact expressible by a single second-order sentence. This axiom gives "local replacement" - in particular, it implies that $V_\alpha$ exists for each ordinal $\alpha$ in the model.

• Finally, add "There is no ordinal which is a limit of inaccessibles."

This characterizes $V_\mu$ up to isomorphism.

$^*$Or rather, the union of the first $\omega$-many $V_\kappa$s with $\kappa$ inaccessible is so characterizable; in general, if $(\kappa_i)_{i\in\omega}$ is an increasing sequence of inaccessibles there is of course no reason to believe that $\bigcup_{i\in\omega} V_{\kappa_i}=V_{\sup\{\kappa_i:i\in\omega\}}$ is so characterizable.

Answering your more specific question: yes, it can be so characterized, in the following way:

• First, start with the axioms of second-order Zermelo set theory with choice (without Replacement). Note that $V_\kappa$ is a model of this if $\kappa$ is the $\omega$th inaccessible - indeed, as long as $\lambda$ is a limit ordinal greater than $\omega$ we have that $V_\lambda$ satisfies second-order Zermelo set theory with choice.

• Next, add "For every ordinal $\alpha$ there is an inaccessible cardinal $>\alpha$."

• Finally, add "There is ordinal which is a limit of inaccessibles."

This characterizes $V_\kappa$ up to isomorphism.

Answering your more specific question: yes, it$^*$ can be so characterized, in the following way:

• First, start with the axioms of second-order Zermelo set theory with choice (without Replacement). Note that $V_\mu$ is a model of this if $\mu$ is the first limit of inaccessibles - indeed, as long as $\lambda$ is a limit ordinal greater than $\omega$ we have that $V_\lambda$ satisfies second-order Zermelo set theory with choice.

• Next, add "For every ordinal $\alpha$ there is an inaccessible cardinal $>\alpha$."

• Finally, add "There is no ordinal which is a limit of inaccessibles."

This characterizes $V_\mu$ up to isomorphism.

$^*$Or rather, the union of the first $\omega$-many $V_\kappa$s with $\kappa$ inaccessible is so characterizable; in general, if $(\kappa_i)_{i\in\omega}$ is an increasing sequence of inaccessibles there is of course no reason to believe that $\bigcup_{i\in\omega} V_{\kappa_i}=V_{\sup\{\kappa_i:i\in\omega\}}$ is so characterizable.