Here's a proof, which is really an expansion of my comment above.
Let $A=\langle a\rangle$ be a cyclic group of order $n$, with $n$ odd. Let $K=\operatorname{Aut}(A)$, and consider the holomorph $G=A\rtimes K$. Because $n$ is odd, $Z(G)$ is trivial: because the inversion map is in $K$, we see that $C_G(a)=A$, and no element of $A$ is central. Also, since $K$ is abelian, we see $A=[G,G]$.
Now let $\phi$ be an automorphism of $G$. Then $\phi(A)=A$, and so there exists $k\in K$ such that $\phi(a)=a^k$. Now $H=\phi(K)$ is a self-centralizing subgroup of $G$ that is a complement to $A$. Because it's a complement, for every $g\in K$, there's a unique element of the form $a^?g\in H$. In particular, consider $\iota\in K$, the inversion map. If $a^r\iota\in H$, then setting $m=-r(n+1)/2$, it is easy to check that $\iota\in a^mHa^{-m}$. By looking at $[\iota,ga^s]$, we see that $C_G(\iota)=K$, and so $a^mHa^{-m}=K$. Thus $\phi$ acts on $G$ exactly like conjugation by $ka^m$, so $\phi$ is inner. Combined with the triviality of $Z(G)$, we see $G$ is complete.
Edit: I might have glossed over one too many details in the end above. Let $\psi\in\operatorname{Aut}(G)$ be conjugation by $ka^m$, and let $\alpha=\phi\psi^{-1}$. Then we've shown $\alpha$ fixes $A$ pointwise, and $K$ setwise. But then for any $g\in K$, we have
\begin{align}
a^g &= \alpha(a^g)\\
&= a^{\alpha(g)}
\end{align}
and thus $g$ and $\alpha(g)$ are two automorphisms of $A$ with the same action, meaning $\alpha(g)=g$. Thus $\alpha$ fixes $K$ pointwise, and since $G=AK$, $\alpha$ is the identity map.
I'm still not entirely sure about the formulation of the question, but the following is probably a counterexample: The holomorph of $C_9$ can be generated by the normal subgroup $N=\langle (1,2,3,4,5,6,7,8,9)\rangle\cong C_9$ and the group $U=\langle (2,3,5,9,8,6)(4,7)\rangle$ that acts as the full automorphism group of $N$.
Also in the holomorph is the regular subgroup $M=\langle (1,2,6,4,5,9,7,8,3)\rangle$, also isomorphic to $C_9$. As isomorphic regular groups, $N$ and $M$ are conjugate in $S_9$ (namely by $\sigma=(3,6,9)$), but they are not conjugate in the holomorph, and $\sigma$ does not normalize $N$ (and thus cannot induce an automorphism of $N$).
Best Answer
Since $\varphi\lambda_a=\lambda_{\varphi(a)}\varphi$, we have that $\lambda(G)$ and $\operatorname{Aut}(G)$ commute, and hence $\lambda(G)\operatorname{Aut}(G)\le\operatorname{Sym}(G)$. Moreover, $(\lambda_b\varphi)\lambda_a(\lambda_b\varphi)^{-1}=$ $\lambda_b(\varphi\lambda_a\varphi^{-1})\lambda_b^{-1}=$ $\lambda_b\lambda_{\varphi(a)}\lambda_{b^{-1}}=$ $\lambda_{b\varphi(a)b^{-1}}\in\lambda(G)$, whence $\lambda(G)\unlhd\lambda(G)\operatorname{Aut}(G)$. Finally, $\lambda_a\in\operatorname{Aut}(G)$ if and only if $a=e_G$ if and only if $\lambda_a=Id_G$, whence $\lambda(G)\cap\operatorname{Aut}(G)=\{Id_G\}$. Now use the theorem and note that $G\cong\lambda(G)$.