Traffic light control by one switch instead of $n$ There are $n$ three-position switches that control the red/yellow/green position of a single traffic light. Whenever the positions of $\textbf{all}$ switches are changed, the colour of the light changes. Prove that the colour of the light is actually controlled by only one switch.
The case $n=1$ is trivial and the case $n=2$ can be case-bashed in a lengthy way (if I have not missed some cases). But I do not have any general strategy and cannot even reformulate the problem in a sensible mathematical rigorous way.
I was also thinking that this 'three-position' might be attacked with $\mathbb{F}_3$? Or perhaps the $3$ does not really matter (this makes more sense but who knows).
Any help appreciated!

Another way to rephrase. You are given an a function $f:(\mathbb F_3)^n\to \mathbb F_3$, which has the property that for all $x_1,x_2,\dots,x_n, y_1,y_2,\dots,y_n\in \mathbb F_3$ such that $x_1\neq y_1,x_2\neq y_2,\dots,x_n\neq y_n$, it holds that
$$
f(x_1,x_2,\dots,x_n)\neq f(y_1,y_2,\dots,y_n)
$$
The interpretation is that each $x_i$ represents the position of one of the switches, and the output of $f$ is the color of the traffic light. The above conditions means that changing all switches should change the color of the light.
The goal is to prove that $f$ is actually only determined by one argument. In other words, there exists a function $g:\mathbb F_3\to \mathbb F_3$, and an index $i\in \{1,2,\dots,n\}$, such that
$$
f(x_1,x_2,\dots,x_n)=g(x_i)
$$
for all inputs $x_1,x_2,\dots,x_n$. 
 A: Induction on $n$. The result is trivial for $n=1$. Assume it is true for $n-1$.  
Case 1. For some particular setting of the first $n-1$ switches, the output is determined by the setting of switch $n$. Suppose the setting of the first $n-1$ switches is $(a_1,\dots,a_{n-1})$. So wlog the assumption is that $f(a_1,\dots,a_{n-1},i)=i$ (for $i=1,2,3$). [If the outputs for these three settings occur in a different order, we can just relabel the positions of switch $n$]. Let $(b_1,\dots,b_{n-1})$ be a disjoint setting of the first $n-1$ switches, ie $a_j\ne b_j$ for $j=1,2,\dots,n-1$. Then $(a_1,\dots,a_{n-1},2)$ and $(b_1,\dots,b_{n-1},1)$ are disjoint, so $f(b_1,\dots,b_{n-1},1)\ne2=f(a_1,\dots,a_{n-1},2)$. Similarly $f(b_1,\dots,b_{n-1},1)\ne3=f(a_1,\dots,a_{n-1},3)$. Hence $f(b_1,\dots,b_{n-1},1)=1$. Similarly, $f(b_1,\dots,b_{n-1},2)=2$ and $f(b_1,\dots,b_{n-1},3)=3$.
Now we claim that any setting of the first $n-1$ switches can be reached by two such steps. To get from $(a_1,\dots,a_{n-1})$ to $(c_1,\dots,c_{n-1})$, we pick any $b_i$ different from both $a_i$ and $c_i$ (where for some $i$ we may have $a_i=c_i$). Then the first move is from $(a_1,\dots,a_{n-1})$ to $(b_1,\dots,b_{n-1})$, and the second move is from $(b_1,\dots,b_{n-1})$ to $(c_1,\dots,c_{n-1})$. We thus end up with the output being solely determined by the last switch.
Case 2. For any setting $(a_1,\dots,a_{n-1})$ of the first $n-1$ switches we can find two settings of the last switch which give the same output. We define this to be the output $g(a_1,\dots,a_{n-1})$ for the first $n-1$ switches.
Suppose $(a_1,\dots,a_{n-1})$ and $(b_1,\dots,b_{n-1})$ are disjoint. Then we can find $h\ne k\in\{1,2,3\}$ st $g(a_1,\dots,a_{n-1})=f(a_1,\dots,a_{n-1},h)\ne f(b_1,\dots,b_{n-1},k)=g(b_1,\dots,b_{n-1})$. So by induction $g$ is determined by one of the first $n-1$ switches. So wlog we have $g(a_1,\dots,a_{n-1})=a_1$. Hence for any $a_1,\dots,a_{n-1}$ we have $f(a_1,\dots,a_{n-1},i)=a_1$ for two values of $i\in\{1,2,3\}$.
Now pick $b_i,c_i\in\{1,2,3\}$ st $b_i=a_i+1,c_i=a_i+2\bmod 3$ for each $i$. Then for two values of $i$ in $\{1,2,3\}$ we have $f(b_1,\dots,b_{n-1},i)=g(b_1,\dots,b_{n-1})=b_1$. So given any $j\in\{1,2,3\}$ we can find $k\ne j$ with $f(b_1,\dots,b_{n-1},k)=b_1$. But $(a_1,\dots,a_{n01},j)$ and $(b_1,\dots,b_{n-1},k)$ are disjoint, so $f(a_1,\dots,a_{n-1},j)\ne b_1$. Similarly, $f(a_1,\dots,a_{n-1},j)\ne c_1$. So we must have $f(a_1,\dots,a_{n-1},j)=a_1$. That completes the induction. $\Box$
