This is not homework, I am just trying to make something clear for myself.
I am trying to show that $C([0,1])$ and $C([1,2])$ are isometric. Here both spaces are endowed with the standard sup metric. I am stuck on the distance preserving part of this proof. I have already worked out that $f(t)=t+1$ is a bijection between $[0,1]$ and $[1,2].$ I want to claim that $\phi:C([0,1]) \to C([1,2])$ defined $\phi(f)(t)=f(t+1)$ is an isometry. My work so far is:
Direct computation yields that $\phi^{-1}(f)=f(x-1).$ Indeed
$$\phi^{-1}(\phi(f))=\phi^{-1}(f(x+1))=f(x-1+1)=f(x)$$ and
$$\phi(\phi^{-1}(f))=\phi(f(x-1))=f(x+1-1)=f(x)$$
so that $\phi$ is bijective. Now, we know from how we derived the bijection between two closed intervals that the map $t \mapsto t+1$ is a bijection between $[0,1]$ and $[1,2].$ In particular $x \in [0,1]$ if and only if $x+1 \in [1,2].$ This is where I'm stuck. I want to show that
$$\underset{t \in [1,2]}{sup}|\phi(f(t))-\phi(g(t))|=\underset{t \in [0,1]}{sup}|f(t)-g(t)|$$
but the above does not make sense since $\phi(f)(t)=f(t+1)$ cannot take $3$ as an input for example. Is my map just wrong, maybe use $f(x-1)$ or something like that, or am I missing something obvious from a set theoretic perspective? Thank you for the help.