This question is, in a way, a follow-up of this earlier question of mine.

**Background**

Let $A$, $B$ and $F$ be finite groups and let $\alpha: A \to F$ and $\beta: B \to F$ be surjective homomorphisms.

Let $A \times_F B$ denote the fibered product of $A$ and $B$ over $F$, defined as the subgroup of $A\times B$ consisting of those elements $(a,b)$ such that $\alpha(a) = \beta(b)$. It is the categorical pullback of the following diagram $$\begin{matrix} & & A \cr & & \downarrow^\alpha \cr B & \stackrel{\beta}{\longrightarrow} & F \cr \end{matrix}$$

Now let $\tau \in \operatorname{Aut}(F)$ be an automorphism and consider the "twisted" fibered product $$A\times_{(F,\tau)} B = \lbrace (a,b) \in A \times B \mid \alpha(a) = \tau(\beta(b))\rbrace.$$ In other words, it is the pullback of the diagram $$\begin{matrix} & & A \cr & & \downarrow^\alpha \cr B & \stackrel{\tau \circ\beta}{\longrightarrow} & F \cr \end{matrix}$$

It follows from Robin Chapman's answer to my earlier question, that in the case where $A$ and $B$ are cyclic groups, $A\times_{(F,\tau)} B$ and $A\times_F B$ are *abstractly* isomorphic. In other words, the isomorphism type of the fibered product is impervious to twisting by automorphisms of $F$.

This situation is not exclusive to cyclic groups. In fact, in a paper I am writing at the moment, a large number of fibered products of ADE subgroups of $\operatorname{Sp}(1)$ arise and in all cases the fibered products do not see the automorphism of $F$, up to isomorphism. The key observation in all cases is that one can lift the automorphism $\tau$ of $F$ to an automorphism of either $A$ or $B$. This is trivial for inner automorphisms, since they lift via surjections, but $F$ often admits automorphisms which are not inner and they too happen to lift.

Naturally, one is always suspicious that something which can be shown to hold by a case-by-case analysis might in fact follow from some general result. Hence my question:

**Question**

How general is this?

More precisely, let me ask two questions.

(1) Do automorphisms always lift via surjections?

If true, this would explain what I have observed, but I suspect this is not true: although inner automorphisms do indeed lift, normal subgroups (which define surjections) need not be preserved under outer automorphisms. And at any rate, this would perhaps be too strong a result. What I *really* want to know is the answer to this next question:

(2) Are twisted fibered products $A\times_{(F,\tau)} B$ corresponding to different automorphisms $\tau$ always abstractly isomorphic?

Thanks in advance!