Elements that satisfy an equation in an affine Artin group of type $\widetilde{A}_2.$

Question

In order to properly express my question, lets set the notation. Consider the group $$A[\widetilde{A}_2]:=\langle a,b,c\ |\ aba=bab,aca=cac,bcb=cbc\rangle,$$ that is, an Artin group of type $\widetilde{A}_2$, and let $\varphi:A[\widetilde{A}_2]\longrightarrow A[\widetilde{A}_2]$ be the automorphism defined as $$\varphi(a)=b,\quad \varphi(b)=c\quad \textrm{and}\quad \varphi(c)=a.$$ I am looking forward to find every non trivial elements $\alpha\in A[\widetilde{A}_2]$, if there are any, that satisfy the realtion $$\alpha\varphi^2(\alpha^2)=\varphi(\alpha^2)\alpha,$$ where $\varphi^2:=\varphi\circ\varphi$.

I have not done much improvement in this problem, but there are some results that might be helpful:

1. The center of $A[\widetilde{A}_2]$ is trivial.
2. The outer automotphism group of $A[\widetilde{A}_2]$ is isomoprhic to $\mathfrak{S}_3\times\mathbb{Z}/2\mathbb{Z}$, where $\mathfrak{S}_3$ is the symmetric group of order $6$ conform by automorphisms that are the permutations of the generators (for instance, both $\varphi$ and $\varphi^2$) and $\mathbb{Z}/2\mathbb{Z}$ is conform by the automorphism that send the generators to their inverses.
3. The element $\varphi^2(\alpha)\alpha\varphi(\alpha)$ is in the centralizer of $\alpha^2$ in $A[\widetilde{A}_2]$. We can see this by applying $\varphi$ to the equation. Indeed $$\alpha\varphi^2(\alpha^2)=\varphi(\alpha^2)\alpha\xrightarrow[]{\varphi}\varphi(\alpha)\alpha^2=\varphi^2(\alpha^2)\varphi(\alpha)\xrightarrow[]{\varphi}\varphi^2(\alpha)\varphi(\alpha^2)=\alpha^2\varphi^2(\alpha),$$ therefore $$\varphi^2(\alpha)\alpha\underbrace{\varphi(\alpha)\alpha^2}=\varphi^2(\alpha)\underbrace{\alpha\varphi^2(\alpha^2)}\varphi(\alpha)=\underbrace{\varphi^2(\alpha)\varphi(\alpha^2)}\alpha\varphi(\alpha)=\alpha^2\varphi^2(\alpha)\alpha\varphi(\alpha).$$
4. The group $A[\widetilde{A}_2]$ is torsion-free and co-Hopfian.
5. We have that $$\varphi(\alpha)\alpha^2\varphi(\alpha)=\varphi^2(\alpha)\alpha^2\varphi^2(\alpha).$$ Again, it is derived similarly as in the point 3.
6. There is a normal form, called Shortlex Minimal Form, which is explained in the article Artin group of large type are shortlex automatic with regular geodesics, written by D. Holt and S. Rees and published in 2012.
7. $\varphi$ has not trivial fixed points.

I will greatly appreciate any help with the problem, and thanks in advance.

A bit of improvement: I have proven that an element $\alpha\in A[\widetilde{A}_2]$ satisfy that $$\alpha\varphi(\alpha)\alpha=\varphi(\alpha)\alpha\varphi(\alpha)$$ if and only if $\alpha\in\{1,a,b,c,a^{-1},b^{-1},c^{-1}\}$. This is proven from the fact that $\varphi$ has non trivial fixed points and that the group is co-Hopfian. Now, using the result sugested by @kabenyuk, notice that since $\varphi(\alpha)\alpha^2\varphi(\alpha)=\varphi^2(\alpha)\alpha^2\varphi^2(\alpha)$ and $\alpha^2=\varphi^2(\alpha)\varphi(\alpha^2)\varphi^2(\alpha^{-1})$, we have that $$\varphi(\alpha)\varphi^2(\alpha)\varphi(\alpha^2)=\varphi^2(\alpha^2)\varphi(\alpha)\varphi^2(\alpha).$$ If we take a look in this equation, my intuition says that the only element that satisfy this is the identity. Nevertheless, notice that we can write this as $$\varphi^2(\alpha^2)=\varphi(\alpha)\varphi^2(\alpha)\varphi(\alpha^2)\varphi^2(\alpha^{-1})\varphi(\alpha^{-1})=\left(\varphi(\alpha)\varphi^2(\alpha)\varphi(\alpha)\varphi^2(\alpha^{-1})\varphi(\alpha^{-1})\right)^2.$$ Therefore, there exists an element $\beta\in A[\widetilde{A}_2]$ such that $$\varphi^2(\alpha)=\beta\varphi(\alpha)\varphi^2(\alpha)\varphi(\alpha)\varphi^2(\alpha^{-1})\varphi(\alpha^{-1})\beta^{-1},$$ that is $$\varphi^2(\alpha)\beta\varphi(\alpha)\varphi^2(\alpha)=\beta\varphi(\alpha)\varphi^2(\alpha)\varphi(\alpha).$$ Also, we know that $\beta$ is in the centralizer of $\varphi^2(\alpha^2)$, since $$\varphi^2(\alpha^2)=\beta\varphi(\alpha)\varphi^2(\alpha)\varphi(\alpha^2)\varphi^2(\alpha^{-1})\varphi(\alpha^{-1})\beta^{-1}=\beta\varphi^2(\alpha^2)\beta^{-1}.$$