Let $G = (V, E)$ be a finite symmetric (i.e. undirected: $(v, v') \in E \Rightarrow (v', v) \in E$) graph. We say that $\phi: G \to G$ is a graph map if $\phi: V \to V$ is a function and given an edge $(v, v') \in E$ we have $(\phi(v), \phi(v')) \in E$ or $\phi(v) = \phi(v')$, that is: the image of an edge is an edge or a vertex (but not two vertices). In a sense, a graph map does not "tear apart" edges.
I need to know if the following statement is true or false:
"For any finite symmetric graph $G = (V, E)$ and any $a, b \in V$, there is a graph map $\phi: G \to G$ such that $\phi(a) = b$ and $\phi(b) = a$".
Notice that the map $\phi$ doesn't need to be surjective or injective.
Image with an example: In blue, on the left, there is a graph $G$; on the right, we see $G$ upside down. The arrows in red show the image of each vertex by the graph map $\phi$.
Any hint will be appreciated.
Edit: this is the progress I've had so far:
Consider the smallest path from $a$ to $b$ in $G$ (if there's more than one, choose any of them), and see it as a subgraph of $G$, let's say $H = (V', E')$. I believe it is true that $H$ is a path graph, as in https://en.wikipedia.org/wiki/Path_graph, so we can consider $V' = \{v_1, \ldots, v_n\}$, $v_1 = a$ and $v_n = b$, with $(v_i, v_{i+1}) \in E'$, for $i=1, \ldots, n-1$ . If we show that there is a graph map $r: G \to H$ such that $r(v) = v$ for all $v \in V'$ (a ''retraction graph map''), we are done. Indeed, if such $r$ exists, we can consider the composite graph map $\phi = \iota \circ f \circ r: G \to G$, where $\iota: H \to G$ is the inclusion and $f: H \to H$ is defined by $f(v_i) = v_{n+1-i}$.
If $a$ and $b$ belong to distinct connected components of $G$ then we can define the required graph map $\phi$ by putting for each vertex $v$ of $G$, $\phi(v)=b$, provided $v$ belongs to the connected component containing $a$, and $\phi(v)=a$, otherwise.
Otherwise we define the map $\phi: G\to H$ as follows. For each vertices $v,v’$ of $G$ let $d(v,v’)$ be the length of a shortest path from $v$ to $v’$, if $v$ and $v’$ belong to one connected component of $G$, and $d(v,v’)$ be $\infty$, otherwise. Now for each vertex $v\in V$ put $\phi(v)=v_{\min\{d(v,b)+1,n\}}$.