Suppose that $K$ is a field of characteristic $0$, that $L$ is the splitting field over $K$ of some polynomial $f$, and that $M$ is an extension of $L$.
If $\alpha_1, \dots, \alpha_n$ are the roots of $f$ in $L$, then $L = K(\alpha_1, \dots, \alpha_n)$.
Suppose that $\varphi:L \to M$ is a $K$-monomorphism (i.e. it fixes all the elements of $K$) with the additional property that for each root $\alpha_i$ of $f$, its image $\varphi(\alpha_i)$ is also a root of $f$. (I.e. $\forall i, \exists j$ such that $\varphi(\alpha_i) = \alpha_j$.)
How does one show that the image of $L$ under $\varphi$ is in $L$?
IOW, how does one show that $\varphi(z) \in L, \; \forall z \in L$ ?
At the heart of this question is that of the relationship between the generators of an extension (i.e. the roots $\alpha_i$ in this case), and a basis (if any) of said extension over the base field ($K$ in this case). This relationship is a continual source of confusion for me, since "set of generators" and "basis" are similar concepts, but not equivalent. (If the $\alpha_i$ were a basis of $L:K$, instead of the set of generators of $L$ over $K$, then the proof of the desired assertion would be obvious.)
NB: I'm not sure how important the stipulation that $K$ is a field of characteristic $0$ is to this question, but it was part of the setting where I found the question.
You know that $f$ can have at most $n$ roots in any extension. Any monomorphism $\varphi: L \to M$ fixing $K$ is determined by where it takes $\alpha_i$, since $L = K(\alpha_1,\dots,\alpha_n)$. But $\varphi$ takes a root $\alpha_i$ to a root of $f$, by @Hurkyl's comment, and since there are only $n$ roots in any extension, the $n$ roots in $M$ must be $\{\alpha_1,\dots,\alpha_n\} \subset L$. So $\varphi$ must take $\alpha_i$ to some $\alpha_j$, which lies in $L$.
In other words, $\varphi: L \to M$ is going to be an automorphism of $L$ followed by the natural inclusion $L \subset M$.
In other words, $\varphi$ first permutes the roots of $\alpha_i$ somehow, then embeds $L$ into $M$ in the natural way.