As a follow up to my previous answer, I have computed all the automorphism series for groups of order up to 24 within the limits of what I could do with GAP. I will share some of what I found, throughout I will use the notation $(n,d)$ for the group of order $n$ with GAP library id $d$. For large groups not contained in the library, I will use $<r,n>$ to denote a group with minimum generating set of size $r$ and of order $n$, if I know only a bound for $r$, I will use $<\leq r,n>$ to denote that fact. We will say that a group $G$ stabilizes (at $k$) if $Aut^{(k-1)}(G) \not\simeq Aut^{(k)}(G) \simeq Aut^{(k+1)}(G)$. We will call a group $G$ stable if $G \simeq Aut(G)$.

I have verified that all groups of order up to 24 stabilize except for the following groups:

$(16,5)$, $(16,6)$, $(16,7)$, $(16,8)$, $(16,9)$, $(16,10)$, $(16,11)$, $(24,4)$, $(24,6)$, $(24,7)$, $(24,9)$, and $(24,10)$.

Of note is that many of these groups have the same automorphism group, hence their series is identical from $k=1$ onwards. Specifically, $(16,5)$, $(16,6)$, $(16,8)$, $(24,9)$, and $(24,10)$ all have $(16,11)$ as their automorphism group. The groups $(16,7)$, $(16,9)$ have the same automorphism group and so do the groups $(24,4)$, $(24,6)$, and $(24,7)$.

The following is a list of the groups I know to be stable, it is complete only up to order 24, and I give a Structure Description of the ones of order up to 24:

$(1,1) \simeq \mathbb{Z}_1$, $(6,1) \simeq S_3$, $(8,3) \simeq D_8$, $(12,4) \simeq D_{12}$, $(20,3) \simeq \mathbb{Z}_5 \rtimes \mathbb{Z}_4$, $(24,12) \simeq S_4$.

These are still in the library: $(40,12)$, $(42,1)$, $(48,48)$, $(54,6)$, $(110,1)$, $(144,183)$, $(336,208)$, $(384,5678)$, $(432,734)$, $(1152,157849)$.

These are too big to be in the library: $<2,40320> \simeq S_8$, $<4,442368> \simeq Aut^{(6)}((16,3))$.

These results give me two (general) ideas on how to attack this problem: One, analyze those groups for which I haven't been able to determine stabilization to see if I can find anything they have in common and use it show the conjecture is false. Two, Analyze the stable groups to see what causes them to be stable and use that knowledge to (somehow) show that every group must eventually stabilize. Both will likely require a detailed analysis of how $Aut(G)$ arises from $G$, to see how $G$ controls the properties of $Aut(G)$.

Edit:
I now have a complete list of stable groups of order up to 255, their GAP structure descriptions already reveal some very interesting patterns:

$(1,1)\simeq\mathbb{Z}_{1}$

$(6,1)\simeq S_{3}$

$(8,3)\simeq D_{8}$

$(12,4)\simeq D_{12}$

$(20,3)\simeq\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4}$

$(24,12)\simeq S_{4}$

$(40,12)\simeq\mathbb{Z}_{2}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$

$(42,1)\simeq(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$

$(48,48)\simeq\mathbb{Z}_{2}\times S_{4}$

$(54,6)\simeq(\mathbb{Z}_{9}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$

$(84,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2})$

$(110,1)\simeq(\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2}$

$(120,34)\simeq S_{5}$

$(120,36)\simeq S_{3}\times(\mathbb{Z}_{5}\rtimes\mathbb{Z}_{4})$

$(144,182)\simeq((\mathbb{Z}_{3}\times\mathbb{Z}_{3})\rtimes\mathbb{Z}_{8})\rtimes\mathbb{Z}_{2}$

$(144,183)\simeq S_{3}\times S_{4}$

$(156,7)\simeq(\mathbb{Z}_{13}\rtimes\mathbb{Z}_{4})\rtimes\mathbb{Z}_{3}$

$(168,43)\simeq((\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{7})\rtimes\mathbb{Z}_{3}$

$(216,90)\simeq(((\mathbb{Z}_{2}\times\mathbb{Z}_{2})\rtimes\mathbb{Z}_{9})\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$

$(220,7)\simeq\mathbb{Z}_{2}\times((\mathbb{Z}_{11}\rtimes\mathbb{Z}_{5})\rtimes\mathbb{Z}_{2})$

$(240,189)\simeq\mathbb{Z}_{2}\times S_{5}$

$(252,26)\simeq S_{3}\times(\mathbb{Z}_{7}\rtimes\mathbb{Z}_{3})\rtimes\mathbb{Z}_{2}$