Prove: Let $H$ be a group of permutations of the set $X$. Let $K$ be a normal subgroup of $H$ having a unique fixed point $y$, i.e. $\{ y \}$ = $\{x \in X : k(x) = x\ \forall k \in K\}$. Then $h(y) = y$ for all $h$ in $H$.
Proof: Let $H$ be a group of permutations of the set $X$. Let $K$ be a normal subgroup of $H$ having a unique fixed point $y$. Therefore, $\forall k \in K, \forall h \in H, hkh^{-1} \in K$. Since $y$ is the unique fixed point for all $k$ in $K$, and $K$ is normal, $hkh^{-1}$ fixes $y$. Then, $hkh^{-1}(y) = y$, which implies $k(h^{-1}(y)) = h^{-1}(y)$. Then $k$ fixes $h^{-1}(y)$. Since $y$ is a unique fixed point in $X$, $h^{-1}(y) = y$. Therefore, $h(h^{-1}(y)) = h(y)$, and hence $h(y) = y$ for all $h$ in $H$.
So this was my attempt at the proof. Sorry for the repetitive nature of the proof relative to the problem statement, I always start with what I am given. I assumed because normality was given, the property needed utilized and that is where I began my train of thought. However, after convincing myself this proof was valid a peer said this was incorrect. I can not see why. Anyone have a clue to the flaw in logic? Thank you.
Your proof uses a stronger assumption, namely, that $$\forall k\in K\ :\ \{y\}=\{x\in X: k(x)=x\}\,.$$ (By the way, it is so strong that $|X|=1$ is implied, as the identity has to be in $K$, and every point is a fixed point for that.)
But basically the same argument goes through, as $h(y)$ (or $h^{-1}(y)$ if you prefer) will be anyway fixed by each $k\in K$.