This is true for any smooth variety $X$ (i.e. separated scheme of finite type over a field) if and only if $\mathbb{P}(\varphi)^*(\mathcal{O}_{\mathbb{P}(E)}(1))=\mathcal{O}_{\mathbb{P}(E)}(1)$. I don't know of any counterexamples to this equality.

For one direction, note that if this equality doesn't hold, then the automorphism induces a morphism that's different on the Picard group which is a subgroup of the Chow ring. Below, I'll show the converse.

**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 variety and let $\varphi:E\rightarrow E$ be any $X$-automorphism of a vector bundle $E$ over $X$. If $\mathbb{P}(\varphi)^*(\mathcal{O}_{\mathbb{P}(E)}(1))=\mathcal{O}_{\mathbb{P}(E)}(1)$, 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 $L=\mathcal{O}_{\mathbb{P}(E)}(1)$. If $\mathbb{P}(\varphi)^*(L)=L$, then 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 $\mathbb{P}(\varphi)^*(L)=L$. $\square$


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. If $X$ is smooth, then by the projective bundle formula (and the relation between the divisor class group and the Picard group) we have $$\mathrm{Pic}(\mathbb{P}(E))=\pi_X^*\mathrm{Pic}(X)\oplus \mathbb{Z}$$ where $L$ generates the copy of $\mathbb{Z}$. 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 $X$.

**Claim 4**. If $E=\mathcal{O}_X^{\oplus n+1}$, then $\mathbb{P}(\varphi)^*$ is the identity.

*Proof*. The composition of a section $\rho: X\rightarrow X\times \mathbb{P}^n$ with $\varphi$ and the projection $\psi:X\times \mathbb{P}^n\rightarrow \mathbb{P}^n$ shows that $$ (\rho\circ\varphi\circ\psi)^*(\mathcal{O}(1))=M.$$ But we also have that the composition $X\rightarrow \mathbb{P}^n$ factors through a point in $\mathbb{P}^n$ so $M=\mathcal{O}_X$.$\square$