Yes, this is true. Suppose that $\Omega_1(K)=K$, and let $t\in K$ be such that $t^2=1$, but $t\notin Z(K)$. Then $M=\langle t,Z(K)\rangle$ is an elementary abelian subgroup. But $K=\cup_{h\in H}\ M^h$, because $H$ acts transitively on $K/\Phi(K)$, and thus every element of $K$ has order $2$, which is absurd.

Note that the above shows more; namely, in your case $\Omega_1(K)=Z(K)$.

It is key that you have this group $H$ of automorphisms. For example, there is a group $L$ of order $64$ such that $Z(L)=\Phi(L)=L'$ is elementary abelian of order $8$, and $\Omega_1(L)=L$.

There are lots of things one can say about $\Omega_1(G)$ for p-groups $G$. It actually doesn't seem that rare that $G=\Omega_1(G)$, though I believe it is slightly more restrictive in the $p=2$ case. Such a group is called *one-stepped*, after Ito, and in the 2-group case this means there is a collection of involutions $t_1,\ldots,t_n$ - with $|G|=2^n$ - such that $\langle t_1,\ldots,t_i\rangle$ has order $2^i$.
A very good reference for all of this is the two-volume *Groups of Prime Power Order* by Berkovich and Janko.