You are correct that the only normal subgroups of $G^n$ are as you describe when $G$ is simple and nonabelian:
Proposition. If $G$ is nonabelian and simple, $n\geq 1$, and $N\triangleleft G^n$, then $N=H_1\times\cdots \times H_n$, where $H_i=\{1\}$ or $H_i=G$ for each $i$.
Proof. The projection of $N$ onto the $n$th component is normal in $G$, hence is either $\{1\}$ or all of $G$. Since $N$ is a subgroup of $\pi_1(N)\times\cdots\times \pi_n(N)$, we may assume that all projections are equal to $G$, and prove that $N=G^n$.
Let $g\in G$, and let $(g,g_2,\ldots,g_n)\in N$ be an element with first coordinate equal to $g$. For every $x\in G$ we have $[(g,g_2,\ldots,g_n),(x,1,\ldots,1)] = ([g,x],1,\ldots,1)\in N;$ thus, $[G,G]\times\{1\}\times\cdots\times\{1\}\subseteq N$; since $[G,G]=G$, it follows that $N$ contains $G\times\{1\}\times\cdots\times \{1\}$. Similar arguments show $N$ contain the $i$th copy of $G$ for each $i$, so $N=G\times\cdots\times G$. $\Box$
Now, for the problem you have, you want to look at $(A_5\times A_5)\rtimes C_2$. If $N$ is normal, then $N\cap (A_5\times A_5) \triangleleft A_5\times A_5$, so the intersection is either trivial, equal to $A_5\times\{1\}$, to $\{1\}\times A_5$, or $N$ contains $A_5\times A_5$. In the latter case, $N=A_5\times A_5$ or $N=(A_5\times A_5)\rtimes C_2$.
If $A_5\times\{1\}\subseteq N$, then for every $g\in A_5$ we have $((1,1),1)((g,1),0)((1,1),1) =((1,g),1)\in N$ hence we would have $A_5\times A_5\subseteq N$; symmetrically if $\{1\}\times A_5\subseteq N$.
So we are left with the case in which $N\cap (A_5\times A_5)=\{1\}$.
If $((x,y),1)\in N$, then $((x^{-1},y^{-1}),0)((x,y),1)((x,y),0) = ((1,1),1)((x,y),1) = ((y,x),1)\in N$ and therefore, $((x,y),1)((y,x),1) = ((x^2,y^2),0)\in N$. Therefore, $x^2=y^2=1$. So any nontrivial element of $N$ must have $A_5\times A_5$ coordinate of exponent $2$. But if $((x,y),1)\in N$ with $x\neq 1$, then there exists $z\in A_5$ such that $xz$ is not of order $2$; then $((z,1),0)((x,y),1)((z^{-1},1),0) = ((zx,y),1)((z^{-1},1),0) = ((xz,yz^{-1}),1)\in N$ which contradicts our computations above. Thus, no element of $N$ can have nontrivial $A_5\times A_5$ component. But then $N=\{1\}$ or $N=C_2$, and $C_2$ is not normal in $(A_5\times A_5)\rtimes C_2$. Thus, $N$ is trivial.
So it seems to me that the only three normal subgroups of $(A_5\times A_5)\rtimes C_2$ (with the nontrivial action) are the trivial subgroup, $A_5\times A_5$, and the whole group.