Let $F$ be a field and $\Gamma$ be an indexing set (possibly infinite). Let $K$ be another field. There is an isomorphism $\sigma:F\longrightarrow K$. Let $\lbrace E_i\rbrace_{i\in\Gamma}$ be a collection of subfields of $\overline{F}$ containing $F$ and $\lbrace L_i\rbrace_{i\in\Gamma}$ be a collection of subfields of $\overline{K}$ containing $K$. There is an isomorphism $\sigma_i:E_i\longrightarrow L_i$ such that the restriction of $\sigma_i$ to $F$ is $\sigma$. Let $E\subseteq \overline{F}$ be the compositum of the $E_i's$ and $L\subseteq\overline{K}$ be the compositum of the $L_i's$.
Can we define an isomorphism $\phi:E\longrightarrow L$ such that the restriction of $\phi$ to $E_i$ is $\sigma_i$ ?
Not always. You will need the following condition too:
If $i,j \in \Gamma$, then $\sigma_i = \sigma_j$ on the restriction $E_i\cap E_j$.
This is clearly a necessary condition and it is easy to construct $E_i,L_i,\Gamma$ such that this condition is violated but the restriction to $F$ is equal. Simply take $F = K = \mathbb Q$ so there is a unique $\sigma$. Take $E_1 = L_1 = \Bbb Q[i,\sqrt 2], E_2 = L_2 = \Bbb Q[i,\sqrt 3]$, $\sigma_1(i) = i, \sigma_1(\sqrt 2)= \sqrt 2$, $\sigma_2(i) = -i, \sigma_2(\sqrt 3)= \sqrt 3$.