We can prove that on finite fields, two irreducible polynomials $f,g$ of the same degree are multiples of each other up to a polynomial change of variables, i.e. there exists $\lambda(t),p(t) \in \Bbb F_q[t]$ such that $f(p(t)) = \lambda(t) g(t)$.
My question is to know what happens on other fields. More precisely:
Can we find two irreducible polynomials $f,g \in \Bbb Q[X]$ of the same degree $≥2$, such that there exists $\lambda(t),p(t) \in \Bbb Q[X]$ such that $f(p(t)) = \lambda(t) g(t)$.
Can we find two irreducible polynomials $f,g \in \Bbb Q[X]$ of the same degree $≥2$, such that there doesn't exist $\lambda(t),p(t) \in \Bbb Q[X]$ such that $f(p(t)) = \lambda(t) g(t)$.
For 1. I started with $f(t)=t^2+1$ and tried to see if I can find a polynomial $p$ such that $f(p(t))$ has an irreducible factor of degree $2$.
For 2. I tried to prove that there is no polynomial $p$ in $\bf Q$ such that $p(x)^2+1$ is a multiple of $x^2+2$, without success.
Thank you for your help!
For part 1 the answer is affirmative. Your choice $f(t)=t^2+1$ works, because $$g(t):=f(t+1)=t^2+2t+2$$ is also irreducible. In other words $p(t)=t+1$, $\lambda(t)=1$. A linear substitution will not turn an irreducible polynomial into a reducible one, so this generalizes (but this way of producing examples is unfortunately not very interesting, and I fully expect you to edit the question).
For part 2 the answer is also affirmative. Your choices $f(t)=t^2+1$ and $g(t)=t^2+2$ work. If $f(p(t))$ is a multiple of $t^2+2$, then $f(p(i\sqrt2))=0$, so we must have $p(i\sqrt2)=\pm i$. If we collect the odd and even degree terms of $p(t)$ together, and write $$ p(t)=p_0(t^2)+t p_1(t^2) $$ for some polynomials $p_0,p_1\in\Bbb{Q}[t]$, then $$ p(i\sqrt2)=p_0(-2)+i\sqrt2p_1(-2). $$ Here $p_0(-2)$ and $p_1(-2)$ are both real numbers. Rational actually. For us to have $p(i\sqrt2)=\pm i$, we thus must have $p_0(-2)=0$ and $p_1(-2)=\pm1/\sqrt2$. The latter requirement cannot be fulfilled for $p_1(-2)$ is rational but $\sqrt2$ is not.