Prove that $H=\{\sigma(n)=n\}$ is not a normal subgroup of $S_n$, $n\ge4$
My attempt: If $H$ is a normal subgroup of $G $ , then $gHg^{-1} \in H$.Now, let $g \in S_n$ such that $g(n) \ne n$. Then, let $\sigma \in H$ and let $\sigma$ can be of the form $(a_1 a_2...a_k)....(a_{k+1} .... a_{m})$ where none of the elements is $n$.Now, $g \sigma g^{-1}=(g(a_1)....g(a_k))....((g(a_{k+1})....g(a_m))$.
Now a possible reason which I was able to come up with is that suppose some $g(a_i)=n$ then to be a cycle $g(n)$ must be in the cycle which is not possible as $\sigma(n)=n$.As conjugation preserves the cycle structure , it is not possible...
Is this okay?
There are some problems with your attempt: Note that $gHg^{-1}$ is a subset of $S_n$, so the expression $$gHg^{-1}\in H,$$ makes no sense. It is true that if $H$ is normal, then $gHg^{-1}\subset H$, and even $gHg^{-1}=H$.
Furthermore, if $g(n)\neq n$ then necessarly $g(a_i)=n$ for some $a_i\in\{1,\ldots,n-1\}$. However, if $\sigma(a_i)=a_i$ then still $g\sigma g^{-1}\in H$, so you haven't shown that $H$ isn't normal.
To show that $H$ isn't normal, pick specific $\sigma\in H$ and $g\notin H$ and verify that $g\sigma g^{-1}\notin H$. Then it follows that $H$ is not normal. For example, take $\sigma=(1\ 2)$ and $g=(1\ n)$ so that $$g\sigma g^{-1}=(1\ n)(1\ 2)(1\ n)=(2\ n).$$