Solved with megarnie. Let $P'$ denote the formal derivative of $P$. Let $F$ be a field of characteristic $0$ and $\overline{F}$ its algebraic closure. Lemma 1. $P\in F[x]$ is separable if and only if $\gcd(P,P')=1$. Proof. Suppose $P(x)$ has a linear factor $x-\alpha$ in $\overline{F}[x]$ and $P(x)=(x-\alpha)Q(x)$ for some $Q\in\overline{F}[x]$. By the product rule, $P'(x)=Q(x)+(x-\alpha)Q'(x)$ so $x-\alpha\mid Q(x)$ if and only if $x-\alpha\mid P'(x)$. The conclusion follows. $\square$ For an algebraic number $\alpha$ over $F$, let $m_\alpha$ denote its minimal polynomial. Lemma 2. If $p\in F[x]$ is irreducible then it is separable. Proof. Note that $p'$ is nonzero and $\deg p'<\deg p$ so $d:=\gcd(p,p')\in F[x]$ has degree less than $\deg p$. If $p$ is not separable, by Lemma 1 it follows that $d$ has degree at least $1$. But $d\mid p$, contradicting the irreducibility of $p$. $\square$ Lemma 3. $P\in F[x]$ is squarefree if and only if it is separable. Proof. Clearly $P$ is squarefree in $F[x]$ if it is squarefree in $\overline{F}[x]$. Conversely, the irreducible factors $\pi_1,\ldots,\pi_n$ of $P$ are precisely minimal polynomials of algebraic elements over $F$. The $\pi_i$ are separable by Lemma 2. If $\pi_i$ shares a linear factor $x-\alpha$ with $\pi_j$ then $\pi_i=\pi_j=m_\alpha$ (up to multiplication by a unit). Thus $P$ is separable if it is squarefree in $F[x]$. $\square$ Corollary 4. $P\in F[x]$ is squarefree if and only if $\gcd(P,P')=1$. We claim the only functions are $\boxed{f(P)\equiv cP'+PQ}$ for $c\in\mathbb{Q}^\times$ and $Q\in\mathbb{Q}[x]$. By Corollary 4, $\gcd(P,cP'+PQ)=\gcd(P,P')=1$ if and only if $P\in\mathbb{Q}[x]$ is squarefree so these functions work. Note that $f$ is linear. We prove a more general statement: generaliation wrote: Let $F$ be a field of characteristic $0$. Let $f:F[x]\rightarrow F[x]$ be linear and satisfy $\gcd(P,f(P))=1$ if and only if $P\in F[x]$ is squarefree. Then $f(P)\equiv cP'+PQ$ for some $c\in F^\times$ and $Q\in F[x]$. Let $P_a:=f(x^a)$. Claim 5. $P_a(x)=ax^{a-1}P_1(x)-(a-1)x^a P_0(x)=(x^a)'(P_1(x)-xP_0(x))+x^aP_0(x)$ Proof. We apply induction on $a$ with the base case trivial. Assume the statement for $a\leq k$. Then \begin{align*} f((x-r)^{k+1})&=\sum_{n=0}^{k+1}(-r)^{k+1-n}\binom{k+1}{n}P_n(x)\\ &=P_{k+1}(x)+\sum_{n=0}^k(-r)^{k+1-n}\left(\binom{k}{n}+\binom{k}{n-1}\right)P_n(x)\\ &=P_{k+1}(x)-rf((x-r)^k)+(-r)^k\sum_{n=0}^{k-1}\left(-\frac{1}{r}\right)^n\binom{k}{n}P_{n+1}(x)\\ \begin{split} &=P_{k+1}(x)-rf((x-r)^k)\\ &\quad+(-r)^k\left((P_1(x)-xP_0(x))\sum_{n=0}^{k-1}\left(-\frac{1}{r}\right)^n\binom{k}{n}(x^{n+1})'+xP_0(x)\sum_{n=0}^{k-1}\binom{k}{n}\left(-\frac{x}{r}\right)^n\right)\\ &=P_{k+1}(x)-rf((x-r)^k)\\ &\quad+(-r)^k\left((P_1(x)-xP_0(x))\left[x\left(1-\frac{x}{r}\right)^k\right]'+xP_0(x)\left(1-\frac{x}{r}\right)^k\right)\\ &\quad-(k+1)x^k(P_1(x)-xP_0(x))-x^{k+1}P_0(x). \end{split} \end{align*}Since $(x-r)^{k+1}$ is not squarefree, $x-r$ divides $f((x-r)^{k+1})$. If $k\geq 2$ then $x-r$ also divides $f((x-r)^k)$ and $\left[x\left(1-\frac{x}{r}\right)^k\right]'$. Thus $x-r$ divides \[ P_{k+1}(x)-(k+1)x^k(P_1(x)-xP_0(x))-x^{k+1}P_0(x) \]for all $r\in F$. Since $F$ is infinite, it follows that \[ P_{k+1}(x)=(k+1)x^k(P_1(x)-xP_0(x))+x^{k+1}P_0(x), \]as desired. If $k=1$ then $x-r$ divides \[ P_2(x)-2xP_1(x)+x^2P_0(x) \]for all $r\in F$ so $P_2(x)=2xP_1(x)-x^2P_0(x)$, as desired. $\square$ Claim 6. $P_1(x)-xP_0(x)=c$ for some $c\in F^\times$. Proof. Assume for the sake of contradiction $P_1(x)-xP_0(x)$ is not a nonzero constant polynomial. Clearly it cannot be the $0$ polynomial as otherwise $x$ and $P_1(x)$ are not relatively prime ($x$ is squarefree). Let $\alpha\in\overline{F}$ be a root of $P_1(x)-xP_0(x)$. By Claim 5, \[ f(m_\alpha)=m_\alpha'(x)(P_1(x)-xP_0(x))+m_\alpha(x)P_0(x). \]Since $m_\alpha$ is irreducible, it is squarefree. Thus $m_\alpha\nmid f(m_\alpha)$ so $x-\alpha\nmid f(m_\alpha)$. But we have \[ m'_\alpha(\alpha)(P_1(\alpha)-\alpha P_0(\alpha))+m_\alpha(\alpha)P_0(\alpha)=0 \]by the definition of $\alpha$, a contradiction. $\square$ By Claim 5 and Claim 6, $P_a(x)=c(x^a)'+x^a P_0(x)$ for some $c\in F^\times$ so $f(P)=cP'+PP_0$ for all $P\in F[x]$, as desired. $\square$

Nice problem. Let $P_n(t)=f(x^n)(t)$. Note that additivity implies that $f$ is $\mathbb{Q}$-linear. Since $t-a \mid f((x-a)^n)$, putting $t=a$ and using linearity, we get $$\sum_{i=0}^n (-1)^{n-i} \binom{n}{i}a^{n-i} P_i(a)=0$$for all rationals $a$. Thus it holds identically, and using induction we can prove: Claim: $P_n(t)=nt^{n-1}P_1(t)-(n-1)t^nP_0(t)$ for all $n \geq 0$. Now, if $P_1(\alpha)=\alpha P_0(\alpha)$ for some complex number $\alpha$, then from the Claim we get $P_n(\alpha)=\alpha^n P_0(\alpha)$. Hence, if $Q$ is the irreducibly rational polynomial of $\alpha$, we get $f(Q)(\alpha)=Q(\alpha)P_0(\alpha)=0$, which is impossible since $Q$ is squarefree, so $Q$ and $f(Q)$ cannot have a common zero $\alpha$. Thus $P_1(t)-tP_0(t)$ has no complex zeros, so $P_1(t)=tP_0(t)+c$ for some $c \neq 0$. This gives $P_n(t)=t^nP_0(t)+cnt^{n-1}$. Hence linearity gives $$\boxed{f(Q)=P_0Q+cQ' \ \ \ \forall Q \in \mathbb{Q}[x]}$$which is indeed a solution whatever $P_0 \in \mathbb{Q}[x]$ and $c \in \mathbb{Q} \setminus \{0\}$ are.

Accidentally wrote this problem oops. Me wrote: Suppose that $\theta$ is a function on integer polynomials such that if $r$ is a double or more of $P$ then $r$ is also a root of $\theta(P)$ $\theta(P) + \theta(Q) = \theta(P + Q)$ Find all possible $\theta$.

To prove the original problem we can note that if $b$ isn't a constant polynomial then taking $P = b$ gives a counter example, and that $b \ne 0$, so the solutions are of the form $aP + bP'$ for polynomial $a$ and nonzero constant $b$.