Is $A_5\times A_5$ the only nontrivial normal subgroup of $(A_5 \times A_5) \rtimes C_2$? The only nontrivial normal subgroups of $A_5 \times A_5$ are $A_5 \times 1$ and $1\times A_5$.  
What are the normal subgroups of $(A_5 \times A_5) \rtimes C_2$?  Is 
$A_5 \times A_5$ the only (nontrivial) one?
In general, if $G$ is a nonabelian simple group, I think the only normal subgroups of $G^n$ are the subgroups of the form $H_1 \times \cdots \times H_n$, where each $H_i$ is either $1$ or $G$.  What are the normal subgroups of $G^n \rtimes S_n$?

1st Update: I just asked GAP and apparently there are four non-trivial normal subgroups of $(A_5 \times A_5) \rtimes C_2$.  Besides the obvious one, $A_5\times A_5$, there are two isomorphic to $A_5$, and one isomorphic to $S_5$.
(See GAP code below.) (wrong!) 
2nd Update: As SteveD noticed, there was a bug in my GAP code, so I've deleted it.  The group I had chosen for $C_2$ actually normalized the second $A_5$ factor, so the resulting group was isomorphic to $S_5\times A_5$, instead of $(A_5 \times A_5) \rtimes C_2$, which explains why I found the normal subgroups mentioned above.  (SteveD provides alternative GAP code in his answer below.)
 A: 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.
A: If your action of $C_2$ is non-trivial, there are only the obvious three: trivial subgroup $\lbrace 1\rbrace$; whole group $G$; and the $A_5\times A_5$ subgroup.
To see this, let $H=A_5\times A_5$ and $K$ be a copy of the $C_2$. If $N$ is a normal subgroup of your group, then $N\cap H$ is a normal subgroup of both $H$ and $G$. Because it is normal in $H$ - and $H$ is semisimple - $N\cap H$ is either all of $H$; trivial; or one of the $A_5$ factors.  If it's all of $H$, that case is easy.  If it's trivial, then $N$ is a conjugate of $K$, and that can't happen. If it's just one $A_5$ factor, it won't be normalized by $K$, contradicting that $N\cap H$ is normal in $G$.
I am unclear on what GAP code you used, but the following works for me:
a:=AlternatingGroup(5);
k:=CyclicGroup(2);
g:=WreathProduct(a,k);
NormalSubgroups(g);

