Assume that $V$ is a vertex operator algebra, and the VOA $V'$ is a vertex subalgebra of $V$. The notion that $V'\subset V$ is a conformal inclusion has different meanings in different literatures. Physicists often mean that $V'$ and $V$ have the same central charge, while mathematicians often mean that $V'$ and $V$ have the same conformal vector. Clearly the second meaning implies the first one. Is the converse also true? i.e., if $V'$ and $V$ have the same central charge, is the conformal vector of $V'$ also a conformal vector of $V$? (Feel free to add conditions on $V$ and $V'$, e.g., CFT type, rational, etc.)
The answer is no: it is possible to have a vertex algebra map between vertex operator algebras of equal central charge that does not take preserve the distinguished conformal vectors.
Given a vertex algebra $V$, the set of conformal vectors of central charge $c$ in $V$ is in natural bijection with the set of vertex algebra homomorphisms from the Virasoro vertex algebra at central charge $c$ to $V$. It is quite easy to find vertex algebras with more than one conformal vector of central charge $c$. For example, any framed vertex operator algebra with central charge at least $1$ has at least 2 Ising vectors, giving different conformal structures at central charge 1/2.
In this case, it is easy to set $V'$ to be $V$ as a vertex algebra, but choose different Ising vectors to be the conformal vectors. Then the identity map is a conformal embedding in the first sense, but not the second.