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As was implicit in my comment every transitive permutation group contains such a subgroup. I single out the case that $\Omega$ has prime power cardinality in my comment ( in which case any minimal transitive subgroup of ${\rm Sym}(\Omega)$ is a $p$-group, using Sylow's theorem). Just one further remark: in the case $|\Omega| = p^{n}$ for a prime $p$ and positive integer $n$, a transitive $p$-subgroup $G$ of ${\rm Sym}(\Omega)$ is a minimal transitive subgroup if and only if each point-stabilizer $G_{\omega}$ is contained in the Frattini subgroup $\Phi(G)$.

For if $G$ has a transitive proper subgroup $H$, then we have $HG_{\omega} = G$, so certainly $G_{\omega} \not \leq \Phi(G)$ as $G = \langle H, G_{\omega} \rangle \neq H$.

On the other hand, if $G$ is minimal transitive, and $M$ is a maximal subgroup of $G$, then $MG_{\omega} \neq G$ (otherwise $M$ would be transitive), so that $MG_{\omega} = M$ by maximality ( for $M \lhd G$, so that $MG_{\omega}$ is a subgroup of $G$). Hence $G_{\omega} \leq M$, so that $G_{\omega} \leq \Phi(G),$ as $M$ was an arbitrary maximal subgroup of $G$.

As was implicit in my comment every transitive permutation group contains such a subgroup. I single out the case that $\Omega$ has prime power cardinality in my comment ( in which case any minimal transitive subgroup of ${\rm Sym}(\Omega)$ is a $p$-group, using Sylow's theorem). Just one further remark: in the case $|\Omega| = p^{n}$ for a prime $p$ and positive integer $n$, a transitive $p$-subgroup $G$ of ${\rm Sym}(\Omega)$ is a minimal transitive subgroup if and only if each point-stabilizer $G_{\omega}$ is contained in the Frattini subgroup $\Phi(G)$.

For if $G$ has a transitive proper subgroup $H$, then we have $HG_{\omega} = G$, so certainly $G_{\omega} \not \leq \Phi(G)$ as $G = \langle H, G_{\omega} \rangle \neq H$.

On the other hand, if $G$ is minimal transitive, and $M$ is a maximal subgroup of $G$, then $MG_{\omega} \neq G$ (otherwise $M$ would be transitive), so that $MG_{\omega} = M$ by maximality ( for $M \lhd G$, so that $MG_{\omega}$ is a subgroup of $G$). Hence $G_{\omega} \leq M$, so that $G_{\omega} \leq \Phi(G),$ as $M$ was an arbitrary maximal subgroup of $G$.

Later note: notice then that when $G$ is a finite $p$-group, there is a bijection between faithful transitive permutation representations of $G$ in which no proper subgroup is transitive, and $G$-conjugacy classes of subgroups $X \leq \Phi(G)$ such that $X \cap Z(G) = 1.$

For if $X$ is such a subgroup, then ${\rm core}_{G}(X) = 1$ ( otherwise, the core meets $Z(G)$ non-trivially), so the (transitive) permutation action of $G$ on the (say right) cosets of $X$ in $G$ is faithful. Each point stabilizer in this action is a $G$-conjugate of $X$, so is contained in $\Phi(G) \lhd G$ as $X \lhd G$. Hence by the above remark, $G$ is a minimal transitive permutation group on the cosets of $X$.

On the other hand, if $G$ has a faithful minimal transitive action on the right cosets of its subgroup $Y$, then $Y \leq \Phi(G)$ as $Y$ is a point stabilizer. Also $Y$ is core-free in $G$ since the permutation action is faithful, so we certainly have $Y \cap Z(G) = 1$.

As was implicit in my comment every transitive permutation group contains such a subgroup. I single out the case that $\Omega$ has prime power cardinality in my comment ( in which case any minimal transitive subgroup of ${\rm Sym}(\Omega)$ is a $p$-group, using Sylow's theorem. Just one further remark: in the case $|\Omega| = p^{n}$ for a prime $p$ and positive integer $n$, a transitive $p$-subgroup $G$ of ${\rm Sym}(\Omega)$ is a minimal transitive subgroup if and only if each point-stabilizer $G_{\omega}$ is contained in the Frattini subgroup $\Phi(G)$.

For if $G$ has a transitive proper subgroup $H$, then we have $HG_{\omega} = G$, so certainly $G_{\omega} \not \leq \Phi(G)$ as $G = \langle H, G_{\omega} \rangle \neq H$.

On the other hand, if $G$ is minimal transitive, and $M$ is a maximal subgroup of $G$, then $MG_{\omega} \neq G$ (otherwise $M$ would be transitive), so that $MG_{\omega} = M$ by maximality ( for $M \lhd G$, so that $MG_{\omega}$ is a subgroup of $G$). Hence $G_{\omega} \leq M$, so that $G_{\omega} \leq \Phi(G),$ as $M$ was an arbitrary maximal subgroup of $G$.

As was implicit in my comment every transitive permutation group contains such a subgroup. I single out the case that $\Omega$ has prime power cardinality in my comment ( in which case any minimal transitive subgroup of ${\rm Sym}(\Omega)$ is a $p$-group, using Sylow's theorem). Just one further remark: in the case $|\Omega| = p^{n}$ for a prime $p$ and positive integer $n$, a transitive $p$-subgroup $G$ of ${\rm Sym}(\Omega)$ is a minimal transitive subgroup if and only if each point-stabilizer $G_{\omega}$ is contained in the Frattini subgroup $\Phi(G)$.

For if $G$ has a transitive proper subgroup $H$, then we have $HG_{\omega} = G$, so certainly $G_{\omega} \not \leq \Phi(G)$ as $G = \langle H, G_{\omega} \rangle \neq H$.

On the other hand, if $G$ is minimal transitive, and $M$ is a maximal subgroup of $G$, then $MG_{\omega} \neq G$ (otherwise $M$ would be transitive), so that $MG_{\omega} = M$ by maximality ( for $M \lhd G$, so that $MG_{\omega}$ is a subgroup of $G$). Hence $G_{\omega} \leq M$, so that $G_{\omega} \leq \Phi(G),$ as $M$ was an arbitrary maximal subgroup of $G$.

See my comment, but for example, a Klein $4$-subgroup of $S_{4}$ which is not the unique normal Klein $4$-subgroup is transitive, but has no transitive proper subgroup.

As was implicit in my comment every transitive permutation group contains such a subgroup. I single out the case that $\Omega$ has prime power cardinality in my comment ( in which case any minimal transitive subgroup of ${\rm Sym}(\Omega)$ is a $p$-group, using Sylow's theorem. Just one further remark: in the case $|\Omega| = p^{n}$ for a prime $p$ and positive integer $n$, a transitive $p$-subgroup $G$ of ${\rm Sym}(\Omega)$ is a minimal transitive subgroup if and only if each point-stabilizer $G_{\omega}$ is contained in the Frattini subgroup $\Phi(G)$.

For if $G$ has a transitive proper subgroup $H$, then we have $HG_{\omega} = G$, so certainly $G_{\omega} \not \leq \Phi(G)$ as $G = \langle H, G_{\omega} \rangle \neq H$.

On the other hand, if $G$ is minimal transitive, and $M$ is a maximal subgroup of $G$, then $MG_{\omega} \neq G$ (otherwise $M$ would be transitive), so that $MG_{\omega} = M$ by maximality ( for $M \lhd G$, so that $MG_{\omega}$ is a subgroup of $G$). Hence $G_{\omega} \leq M$, so that $G_{\omega} \leq \Phi(G),$ as $M$ was an arbitrary maximal subgroup of $G$.