Let $f$ be a monic irreducible polynomial of degree $n$ over $\mathbb{Z}[x]$. This means that $K:=\mathbb{Q}[x]/f(x)$ is a number field.
Let $\theta_1:=\theta,\theta_2,\ldots,\theta_n$ be all roots of $f$, then $K\cong \mathbb{Q}(\theta)$ and all conjugate fields $K_i:=\mathbb{Q}(\theta_i)$, for $2\leq i\leq n$ are also isomorph to $K$. On the other hand, $L:=\mathbb{Q}(\theta_1,\theta_2,\ldots,\theta_n)$ is the splitting field of $f$ over $\mathbb{Q}$.
We also know that $L=\mathbb{Q}(\gamma)$ for a primitive element $\gamma$. Does is mean that $L\cong K\cong K_i$, for $2\leq i\leq n$ through natural automorphisms of type $\gamma\mapsto\theta_i$? Is this the case only for algebraic number fields? Does this imply that all such $K$ are Galois extensions of $\mathbb{Q}$?
If $K_i \cong \mathbb{Q}(\theta_i)$, then you are correct in that $K \cong K_i$ for all $i$. We do not have $L \cong K_i$ in general.
Consider any field $F$. We can view $F$ as a vector space over the smallest subfield of $F$ containing $1$ (call it $F'$). Now suppose we have an isomorphism $\phi: F \rightarrow F_1$.
Claim: $[F:F'] = [F_1:F']$.
Proof: As with any linear map between vector spaces, $\phi$ is determined by its action on a set of basis vectors in $F$, say $\{\alpha_i\}_{i=1}^n$.
Since $\phi$ is injective, it must map the $\alpha_k$'s to a set of linearly independent elements $\{\beta_i\}_{i=1}^n$ in $F_1$. Otherwise, if the $b_k$'s were linearly dependent, we could write $c_1\beta_1 + \cdots + c_n\beta_n = 0$ for $c_k \in F'$ not all zero. By linearity, this would mean $\phi$ maps $c_1\alpha_1 + \cdots + c_n \alpha_n$ to $0$, contradicting injectivity since $0$ itself maps to $0$.
Further, since the $\alpha_k$'s are independent and span $F$, it is clear that the set $\{\beta_i\}$ spans $\text{Image}(\phi) = F_1$ per linearity. Therefore, the dimension of $F$ over $F'$ equals the dimension of $F_1$ over $F'$. In other words, $[F:F']=[F_1:F']$. $\qquad \blacksquare$
In the given scenario, $F' = \mathbb{Q}$. Let's look at a concrete example: $f(x) = x^3 -2$, which is irreducible over $\mathbb{Z}$. It's roots are $\sqrt[3]{2}, \ \omega \sqrt[3]{2}, \ \omega^2 \sqrt[3]{2}$ with $\omega$ a primitive cube root of unity. If we adjoin a single root, say $\mathbb{Q}( \sqrt[3]{2})$, this yields a degree $3$ extension. Notice, however, that this field has no non-real elements, whereas the splitting field of $f$ clearly does. To arrive at a splitting field, we must adjoin at least one more element to $\mathbb{Q}(\sqrt[3]{2})$ -- and the resulting field has degree greater than $1$ over $\mathbb{Q}(\sqrt[3]{2})$. Per multplicativity of degrees, we see that the splitting field must have greater degree over $\mathbb{Q}$ than $\mathbb{Q}(\sqrt[3]{2})$ has over $\mathbb{Q}$.
Due to the claim above, we thus cannot have an isomorphism between the splitting field and $\mathbb{Q}(\sqrt[3]{2})$. Further, since $\mathbb{Q}(\sqrt[3]{2}) \cong \mathbb{Q}(\omega^k \sqrt[3]{2})$, indeed we cannot have an isomorphism between the splitting field and $\mathbb{Q}(\theta_i)$ for any of the roots $\theta_i$.