$\text{Aut}(A_6)$ is not split extension of $A_6\cong \text{Inn} A_6$, i.e. $\text{Aut}(A_6) \not \cong A_6 \rtimes ( \mathbb Z_2 \times \mathbb Z_2)$ Claim:
Short exact sequence $1 \longrightarrow \operatorname{Inn}\left(A_{6}\right) \longrightarrow \operatorname{Aut}\left(A_{6}\right) \longrightarrow \operatorname{Out}\left(A_{6}\right) \longrightarrow 1$ is not right split,
where $\operatorname{Inn}\left(A_{6}\right)\cong A_6$, $\operatorname{Aut}\left(A_{6}\right)\cong\operatorname{Aut}\left(S_{6}\right)\cong S_6\rtimes \mathbb Z_2$ and $\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2$.
To move this question out of the unanswered list, I put my proof in the answer.
Thanks for your time and patience :)
 A: Claim:
Short exact sequence $1 \longrightarrow \operatorname{Inn}\left(A_{6}\right) \longrightarrow \operatorname{Aut}\left(A_{6}\right) \longrightarrow \operatorname{Out}\left(A_{6}\right) \longrightarrow 1$ is not right split,
where $\operatorname{Inn}\left(A_{6}\right)\cong A_6$, $\operatorname{Aut}\left(A_{6}\right)\cong\operatorname{Aut}\left(S_{6}\right)\cong S_6\rtimes \mathbb Z_2$ and $\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2$.
Proof:
$1$. Prerequisites:
(1) Element in $\operatorname{Aut}\left(A_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$ swaps conjugate classes $(abc)$ and $(abc)(def)$ in $A_6$.
(2) Element in $\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ swaps conjugate classes $(ab)$ and $(ab)(cd)(ef)$ in $S_6$,
swaps conjugate classes $(abc)$ and $(abc)(def)$ in $A_6$(also in $S_6$).
$2$. Suppose the sequence right splits and $\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\leqslant\operatorname{Aut}\left(A_{6}\right)$ 
where $\sigma, \rho\in \operatorname{Aut}\left(A_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\langle \sigma\rangle\langle\rho\rangle\cap\operatorname{Inn}(A_6)=1$.
Since $\operatorname{Aut}\left(A_{6}\right)=\operatorname{Aut}\left(S_{6}\right)$, $\rho$ and $\sigma$ can be considered as elements in $\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$.
If $\sigma, \rho\in \operatorname{Inn}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\sigma\rho\in \operatorname{Inn}\left(A_{6}\right)$. Contradiction.
If $\rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ and  $\sigma\in \operatorname{Inn}\left(S_{6}\right)\setminus\operatorname{Inn}\left(A_{6}\right)$, then $\sigma\rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.
$\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\cong\langle \sigma\rho\rangle\langle\rho\rangle$ and $\sigma\rho, \rho\in \operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.
So if the sequence right splits, we can always assume 
$\operatorname{Out}\left(A_{6}\right)\cong\mathbb Z_2\times \mathbb Z_2\cong\langle \sigma\rangle\langle\rho\rangle\leqslant\operatorname{Aut}\left(A_{6}\right)$ where $\rho, \sigma\in\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$.
$3$. $[\operatorname{Aut}\left(S_{6}\right):\operatorname{Inn}\left(S_{6}\right)]=2$ , $\sigma\operatorname{Inn}\left(S_{6}\right)=\rho\operatorname{Inn}\left(S_{6}\right)$, $\rho^{-1}\sigma\in\operatorname{Inn}(S_6)$.
Suppose $\rho^{-1}\sigma=c_\gamma$, where $c_\gamma$ is action of conjugation by $\gamma\in S_6$. 
Since $\langle \sigma\rangle\langle\rho\rangle\cap\operatorname{Inn}(A_6)=1$, $\gamma\in S_6\setminus A_6$ is odd permutation.
$(\rho^{-1}\sigma)^2=c_\gamma^2=1$ gives $\gamma^2=1$, $\gamma$ is transposition or product of three disjoint transpositions.
$\sigma\rho=\rho\sigma$ gives $\rho(\gamma)=\gamma$. 
But $\rho\in\operatorname{Aut}\left(S_{6}\right)\setminus\operatorname{Inn}\left(S_{6}\right)$ swaps conjugate classes $(ab)$ and $(ab)(cd)(ef)$. 
Contradiction. $\Box$
