On some special homomorphisms with conjugate images Let $G$ and $A$ be two finite group. Let $\beta$, $\alpha :G\rightarrow{\rm Aut}(A)$ be two homomorphisms. In general, it is not true that if $\beta$ and $\alpha$ have conjugate images then there exist $\sigma \in{\rm Aut}(A)$ such that $\beta(g)=\sigma \circ \alpha (g)\circ \sigma^{-1}$ for all $g\in G$, but I think that can be hold for a group $G$ in which all its automorphisms are inner. But I don't know how to do this.
Thank you very much.
 A: Let $G$ and $H$ be finite groups (for example, $H=\operatorname{Aut}(A)$) and $\alpha,\beta:G \to H$ two homomorphisms.
Suppose there is some $\sigma \in H$ so that $\beta(g) = \sigma \cdot \alpha(g) \cdot \sigma ^{-1}$ for all $g\in G$. Then the image of $\alpha$ is conjugate (via $\sigma$) to the image of $\beta$. Furthermore, $\ker(\alpha)=\ker(\beta)$ since the only element conjugate to the identity of $H$ is the identity of $H$.
(So you have left out a necessary condition that $\ker(\alpha) = \ker(\beta)$).
Suppose now that $\ker(\alpha) = \ker(\beta)$ so that the kernels are equal, $\operatorname{im}(\beta)^\sigma = \operatorname{im}(\alpha)$ so that images are conjugate, and $G/\ker(\alpha) = G/\ker(\beta)$ has no outer-automorphisms. What does this have to do with $\alpha$ and $\beta$?
Define $\beta'(g) = \sigma^{-1} \cdot \beta(g) \cdot \sigma$ so that $\alpha$ and $\beta'$ have the same image $I$ and the same kernel $K$. Define $\bar \alpha:G/K \to I : gK \mapsto \alpha(g)$ and $\bar \beta'$ similarly.
Then $\bar \alpha$ and $\bar \beta'$ are two isomorphisms from $G/K$ to $I$, so their "difference", $\phi:G/K \to G/K: gK \mapsto \bar\alpha^{-1}(\bar\beta'(gK))$ is an automorphism of $G/K$.
If we assume $G/K$ has no outer-automorphisms, then $\phi$ is inner, so there is some $xK \in G/K$ such that $\phi(gK) = xgx^{-1} K$. Expanding the definition gives $xgx^{-1} K = \bar\alpha^{-1}(\bar\beta'(gK))$ so
$\alpha(x) \alpha(g) \alpha(x)^{-1} = \beta'(g)$ and so taking $\sigma' = \sigma \cdot \alpha(x)$ we get $\beta(g) = \sigma \beta'(g) \sigma^{-1} = \sigma \alpha(x) \alpha(g) \alpha(x)^{-1} \sigma^{-1} = \sigma' \alpha(g) (\sigma')^{-1}$ as requested.
