Proving that two quotient fields of $\mathbb{Q}[x]$ are isomorphic 
Let $f(x)=x^3-3x^2+3x+1, g(x)=x^3+3x^2+3x+3$. I would like to show that $\mathbb{Q}[x]/\left<f(x)\right> \cong \mathbb{Q}[x]/\left<g(x)\right>$.

There is the obvious way: "Let $a$ be a root of $f(x)$, $b$ be a root of $g(x)$, then
the isomorphism is determined by the image of $a$ which must satisfy $\phi(a^3-3a^2+3a+1) = c_1b^2+c_2b+c_3 = 0$", but the calculations needed are a bit long, so is there a quicker but rigorous way to show it by noticing that $f(x)=g(x-2)$ and mentioning that translation by a constant is an automorphism of $\mathbb{Q}[x]$?
 A: Following your observation, the ring homomorphism $\Bbb Q[x]\to \Bbb Q[x]$ given by $x\mapsto x+2$ maps $f\mapsto g$, hence composed with the projection to $\Bbb Q[x]/(g)$ has $f$ in its kernel, hence induces a homomorphism $\Bbb Q[x]/(f)\to\Bbb Q[x]/(g)$. Starting from $x\mapsto x-2$, we obtain an inverse homomorphism, hence we have an isomorphism.
A: Yes, you have exactly the right idea for the quicker proof!
First, for any $h \in \mathbb{Q}[x]$, we can define $\Phi_h : \mathbb{Q}[x] \to \mathbb{Q}[x]$ by $\Phi_h(F) = F \circ h$ (where $\circ$ denotes compositon). You can check directly that $\Phi_h$ is a homomorphism for all $h$. Moreover, $\Phi_x = \operatorname{id}_{\mathbb{Q}[x]}$, and $\Phi_{h_1 \circ h_2} = \Phi_{h_2} \circ \Phi_{h_1}$ for all $h_1, h_2 \in \mathbb{Q}[x]$.
Now notice that $(x-2) \circ (x+2) = (x+2) \circ (x-2) = x$, so $\Phi_{x-2} \circ \Phi_{x+2} = \operatorname{id}_{\mathbb{Q}[x]} = \Phi_{x+2} \circ \Phi_{x-2}$. Thus, $\Phi_{x+2}$ is an automorphism of $\mathbb{Q}[x]$ with inverse $\Phi_{x-2}$. Note that $\Phi_{x+2}(f) = g$ in your example.
In general, if $\alpha : R \to S$ is a ring isomorphism and $I$ is a two-sided ideal of $R$, then $\alpha(I)$ is a two-sided ideal of $S$, and $\alpha$ induces a ring isomorphism $R/I \to S/\alpha(I)$, given by $r + I \mapsto \alpha(r) + \alpha(I)$.
So, we conclude that $\Phi_{x+2}$ induces an isomorphism
$$\mathbb{Q}[x]/\langle f \rangle \to \mathbb{Q}[x]/\Phi_{x+2}(\langle f \rangle) = \mathbb{Q}[x]/\langle g \rangle.$$
