I am trying to show that all group automorphisms of the form $A(x) = g_1xg_2x\cdots x g_n$ for fixed elements $g_1, g_2, \cdots g_n \in G$ are conjugations. It is clear that by setting $x$ to the identity we can conclude that $g_1g_2\cdots g_n$ is the identity. However, I have no idea how to show $n=2$. Any hint will be appreciated.
Edit: As pointed out in the comment, this is not true in Abelian groups. However, I'm trying to show this for a larger proof where G is assumed to be non-Abelian.
Note that your maps are actually "pointwise products" of conjugation maps: if this map is a homomorphism, then as you note we obtain that $g_1\cdots g_n=e$ (by plugging in $x=e$), and therefore $g_n=(g_1\cdots g_{n-1})^{-1}$. Then $$\begin{align*} g_1xg_2x\cdots xg_n &= (g_1)x(g_1^{-1})(g_1g_2)x(g_1g_2)^{-1}(g_1g_2g_3)x \cdots (g_1\cdots g_{n-1})xg_n\\ &=(g_1xg_1^{-1})(g_1g_2)x(g_1g_2)^{-1}(g_1g_2g_3)x \cdots (g_1\cdots g_{n-1})x(g_1\cdots g_{n-1})^{-1}\\ &= \varphi_{g_1}(x) \varphi_{g_1g_2}(x)\cdots \varphi_{g_1\cdots g_{n-1}}(x), \end{align*}$$ where $\varphi_g$ is the inner automorphism determined by $g$, $\varphi_g(x)=gxg^{-1}$.
For $n=3$ (that is, the pointwise product of two automorphisms), we get that this must be an abelian group:
Theorem. Let $G$ be a group, and let $\varphi,\psi\in\mathrm{Aut}(G)$. If $x\longmapsto \varphi(x)\psi(x)$ is an automorphism of $G$, then $G$ is abelian.
Proof. Let $x,y\in G$. Since the map is a homomorphism, we have: $$\begin{align*} \varphi(xy)\psi(xy) &= \varphi(x)\psi(x)\varphi(y)\psi(y)\\ \varphi(x)\varphi(y)\psi(x)\psi(y) &= \varphi(x)\psi(x)\varphi(y)\psi(y)\\ \varphi(y)\psi(x) &= \psi(x)\varphi(y). \end{align*}$$ Fix $x$; as $y$ ranges over all elements of $G$, we get that $\psi(x)$ commutes with all elements of $G$ (since $\varphi$ is surjective. Thus, $\psi(x)\in Z(G)$. Since the center is characteristic, it follows that $x\in Z(G)$. Thus, all elements of $G$ are central, so $G$ is abelian. $\Box$
Corollary. If $A(x) = g_1xg_2xg_3$ is an automorphism, then $G$ is an abelian group with no elements of order $2$ and $A(x)=x^2$. If $G$ is finite, then $G$ is of odd order.
Proof. We have that $A(x) = \varphi_{g_1}(x)\varphi_{g_1g_2}(x)$, so by the THeorem $G$ is abelian, and in particular $\varphi_{y}(x)=x$ for all $y\in G$. Thus, $A(x)=x^2$. As this is an automorphism, $x^2=e$ implies $x=e$. $\Box$
In particular, if $G$ is non-abelian, then $n=3$ is impossible.
However, larger values of $n$ are possible. Specifically, if $G$ has exponent $k$, then you can take $n=k+1$ and $g_1=\cdots g_n=e$. Then $A(x) = x^{n} = x^{k+1}=x$ for all $x\in G$, so $A$ is of course an automorphism. Note that there are nonabelian groups of exponent $n$ for every $n\neq 1,2$.
I'm not sure what happens if we require that the $g_i$ be non-central; you still have a pointwise product of $n-1$ (nontrivial inner) automorphisms. Using the same trick as for $n=3$, we can conclude that we have endomorphisms $\psi_1,\ldots,\psi_k$, and for all $x,y\in G$ we have $$\begin{align*} \psi_1(x)\psi_1(y)\cdots \psi_k(x)\psi_k(y)&= \psi_1(x)\cdots\psi_k(x)\psi_1(y)\cdots\psi_k(y)\\ \psi_1(y)\psi_2(x)\psi_2(y)\cdots\psi_{k-1}(y)\psi_k(x) &= \psi_2(x)\cdots \psi_k(x)\psi_1(y)\cdots\psi_{k-1}(y), \end{align*}$$ but I don't see how one might exploit that.