Can any monomorphism of a subfield of a splitting field be extended to an automorphism? It's a common theorem in field theory that if $\varphi: F\to\overline{F}$ is a field isomorphism, and if $E$ and $\overline{E}$ are splitting fields of monic polynomials $f(x)$ and $\overline{f}(x)$, respectively, where $\overline{f}$ is the image of $f$ under $\varphi$, then $\varphi$ can be extended to an isomorphism of $E$ and $\overline{E}$. 
I'm wondering about the following: Suppose $E$ is a splitting field of a polynomial $f(x)$, and $k$ is a subfield of $E/F$. Can any monomorphism of $\varphi:k\hookrightarrow E$ be extended to a field isomorphism of $E$? Defining $\overline{k}:=\varphi(k)$, I get the case that $\varphi:k\to\overline{k}$ is a field isomorphism, and $f(x)$ and $\overline{f}(x)$ both  split over $E$, when regarded as polynomials in $k$ and $\overline{k}$. But is $E$ still a splitting field of $f(x)$ and $\overline{f}(x)$ as polynomials in $k$ and $\overline{k}$, or is it possibly too big to be a splitting field?
If it is a splitting field, then I suppose it would follow that any monomorphism on a subfield can be extended to an isomorphism of the splitting field.
 A: $E$ is definitely a splitting field for $f(x)$ over $k$, as it is a splitting field over $F \subseteq k$.
Now your statement is clearly true if you assume that $\varphi$ fixes $F$ (pointwise), because then $f(x) = \bar f (x)$. Otherwise, I do not think it is true.
Let me try and describe an example, barring mistakes. 
Let $\alpha = \sqrt[3]{2}$, and $F = \mathbf{Q}[\alpha]$. Let $E$ be the splitting field of $f(x) = x^{3} - \alpha$ over $F$. We have $\lvert E : F \rvert = 6$, and $E$ contains a primitive $3$rd root $\omega$ of unity.
Let $k = F$, and $\varphi(\alpha) = \beta = \omega \alpha \in E$, so that $\bar k = \mathbf{Q}[\beta] \subseteq E$. Here $\bar f(x) = x^{3} - \beta$. But it seems to me that $E$ is no splitting field for $\bar f(x)$. In fact if $a$ is a root of $f(x)$ in $E$, and $b$ is (by way of contradiction) a root of $\bar f(x)$ in $E$, then $(a^{-1} b)^{3} = \alpha^{-1} \beta = \omega$, so that $E$ contains a primitive $9$th root $\theta$ of unity, and $E \supseteq F[\theta]$.
Now note that $E \ne F[\theta]$, for instance because $\operatorname{Gal}(E/F) \cong S_{3}$, while $\operatorname{Gal}(F[\theta]/F)$ is abelian.
It follows that no root of $x^{3} - 2$ is in $\mathbf{Q}[\theta]$, otherwise $E = F[\theta]$. Thus  $$\lvert \mathbf{Q}[\alpha, \theta] : \mathbf{Q} \rvert =\lvert (\mathbf{Q}[\theta])[\alpha] : \mathbf{Q}[\theta] \rvert \cdot \lvert \mathbf{Q}[\theta] : \mathbf{Q} \rvert = 3 \cdot 6 = 18,$$ so that $$\lvert F[\theta] : F \rvert = 
\frac{\lvert (\mathbf{Q}[\alpha])[\theta] : \mathbf{Q} \rvert}{\lvert \mathbf{Q}[\alpha] : \mathbf{Q} \rvert} = \frac{18}{3}
=
6.$$
This implies once more $E = F[\theta]$, a contradiction.
