This is true for any proper variety $X$ (i.e. a proper separated scheme of finite type over a field).

**Claim 1**: Let $X$ be any variety with irreducible components $X_i$. Let $\varphi: E\rightarrow E$ be any $X$-automorphism of a vector bundle $E$ over $X$. Let $\varphi_i$ be the restriction of $\varphi$ to $E|_{X_i}$. If $\mathbb{P}(\varphi_i)^*$ is the identity for all $i$, then $\mathbb{P}(\varphi)^*$ is the identity.

*Proof.* Let $\rho_i:\mathbb{P}(E|_{X_i})\rightarrow \mathbb{P}(E)$ be the inclusion. Then $\rho_{i*}\circ \mathbb{P}(\varphi_i)^*= \mathbb{P}(\varphi^*)\circ \rho_{i*}$ by Proposition 1.7 of Fulton. The result follows since the images of $\rho_{i*}$ generate $\mathrm{CH}(\mathbb{P}(E))$. $\square$


**Claim 2**: Let $X$ be any proper variety and let $\varphi:E\rightarrow E$ be any $X$-automorphism of a vector bundle $E$ over $X$. Then $\mathbb{P}(\varphi)^*$ is the identity.

*Proof.* We'll proceed by Noetherian induction on the dimension of the irreducible components of $X$. For this purpose, let $X$ be any variety with irreducible components $X_i$ having dimension $\max_i\{ \mathrm{dim}(X_i)\}\leq n$.

For the base case, we observe that if $X$ is a point, then any automorphism of $\mathbb{P}^m_X$ has to take $\mathcal{O}(1)$ to itself in $\mathrm{Pic}(X)$ and this generates the ring $\mathrm{CH}^*(\mathbb{P}^m_X)$.

Assume that the result is true for all $X$ with irreducible components $X_i$ with $\max_i\{\mathrm{dim}(X_i)\}\leq n-1$. By Claim 1, it suffices to show that this implies the result when $X$ is irreducible. Then for any open $U\subset X$ there is a commuting ladder with exact rows

$$\begin{matrix} \mathrm{CH}(\mathbb{P}(E|_Z))&\rightarrow& \mathrm{CH}(\mathbb{P}(E))&\rightarrow& \mathrm{CH}(\mathbb{P}(E|_U))&\rightarrow & 0 \\ \downarrow && \downarrow && \downarrow\\  \mathrm{CH}(\mathbb{P}(E|_Z))&\rightarrow& \mathrm{CH}(\mathbb{P}(E))&\rightarrow& \mathrm{CH}(\mathbb{P}(E|_U))&\rightarrow & 0\end{matrix}$$ where the vertical morphisms are induced by $\mathbb{P}(\varphi)^*$. By our induction hypothesis, the left-most arrow is the identity whenever $U$ is nonempty. Taking a limit over the inductive system of closed subschemes $Z\subset X$, we get a commutative ladder with exact rows given by localization
$$\begin{matrix} \varinjlim_Z \mathrm{CH}(\mathbb{P}(E|_Z))&\rightarrow& \mathrm{CH}(\mathbb{P}(E))&\rightarrow& \mathrm{CH}(\mathbb{P}(E|_{\eta}))&\rightarrow & 0 \\ \downarrow && \downarrow && \downarrow\\  \varinjlim_Z \mathrm{CH}(\mathbb{P}(E|_Z))&\rightarrow& \mathrm{CH}(\mathbb{P}(E))&\rightarrow& \mathrm{CH}(\mathbb{P}(E|_{\eta}))&\rightarrow & 0\end{matrix}$$ where $\eta$ is the generic point of $X$. By our induction, the rightmost vertical arrow is the identity. These sequences are (canonically) split by Claim 3 below. A diagram chase finishes the proof. $\square$

**Claim 3**: Let $L=\mathcal{O}_{\mathbb{P}(E)}(1)$. The map $p^*:\mathrm{CH}_i(\mathbb{P}(E))\rightarrow \mathrm{CH}_{i-\mathrm{dim}(X)}(\mathbb{P}(E|_\eta))$ is right-split by the map $s:\mathrm{CH}_{i-\mathrm{dim}(X)}(\mathbb{P}(E|_\eta))\rightarrow \mathrm{CH}_i(\mathbb{P}(E))$ sending $c_1(L|_\eta)^j$ to $c_1(L)^j$. Moreover, there is a commutative square $s\circ\mathbb{P}(\varphi|_\eta)^*=\mathbb{P}(\varphi)^*\circ s$.

*Proof*. Since the group $\mathrm{CH}_i(\mathbb{P}(E|_\eta))=\mathbb{Z}$, for the first claim it suffices to see that $p^*c_1(L)=c_1(L|_\eta)$ which is true by functorality of the pullbacks. To see the second claim, we need to check
$$s\circ\mathbb{P}(\varphi|_\eta)^*(c_1(L|_\eta)^j)=\mathbb{P}(\varphi)^*\circ s(c_1(L|_\eta)^j).$$ The left hand side is equivalent to
$$s\circ\mathbb{P}(\varphi|_\eta)^*(c_1(L|_\eta)^j)=s(c_1(\mathbb{P}(\varphi)^*(L|_\eta))^j)=s(c_1(L|_\eta)^j)=c_1(L)^j$$ while the right hand side is equivalent to $$\mathbb{P}(\varphi)^*\circ s(c_1(L|_\eta)^j)=\mathbb{P}(\varphi)^*(c_1(L)^j))=c_1(\mathbb{P}(\varphi)^*(L))^j$$ by functorality. It suffices then to observe that $\mathbb{P}(\varphi)^*(L)=L$. This is done in Claim 4 below. $\square$

**Claim 4**: Suppose that $E=\mathcal{O}_X^{\oplus n}$. Let $\pi_X:\mathbb{P}(E)\rightarrow X$ be the structure map and let $\mathbb{P}(\varphi):\mathbb{P}(E)\rightarrow \mathbb{P}(E)$ be an $X$-isomorphism. Then $\mathbb{P}(\varphi)^*(L)=L$ where $L=\mathcal{O}_{\mathbb{P}(E)}(1)$.

*Proof*. By [Hart, Chapter III, Section 12, Exercise 5] we have $\mathrm{Pic}(\mathbb{P}(E))=\pi_X^*\mathrm{Pic}(X)\oplus \mathbb{Z}$ where $L$ generates the copy of $\mathbb{Z}$ (technically, this exercise is stated for algebraically closed based fields but this is sufficient for this proof since field extensions of the base induce injections on the Picard group [\[Stacks, Tag 0CC5][1]]). The projection formula then yields 
\begin{align*}\pi_{X*}\mathbb{P}(\varphi)^*(L) &=\pi_{X*}(L\otimes \pi_X^*(M))\\ &= \pi_{X*}(L)\otimes M\\ &=E\otimes M\end{align*} with $M$ a line bundle on $X$.......................... I'll have to come back to it.

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References:

[Hart] Hartshorne - Algebraic Geometry

[Stacks] The Stacks project


  [1]: https://stacks.math.columbia.edu/tag/0CC5