I spoke to my colleague Song Sun, and he reminded me of a discussion that he and I had about Question 2 some time ago. For $n\geq 2$, there are many examples of locally free sheaves on $X_{n} = \mathbb{A}^{n+1}\setminus\{0\}$ that admit no equivariant structure. Denote by $S$ the polynomial ring $\mathbb{C}[x_0,\dots,x_n]$. Denote by $S_+$ the maximal ideal $\langle x_0,\dots,x_n \rangle$. Let $\underline{f}=(f_0,\dots,f_n)$ be a regular sequence of elements in $S_+$.

Associated to this regular sequence there is a Koszul complex $(K_\bullet(\underline{f}),d_\bullet)$ of $S$-modules where $K_0(\underline{f})$ equals $S$, where $K_1(\underline{f})$ equals $S^{\oplus(n+1)}$, where $K_r(\underline{f})$ is the $r^{\text{th}}$ exterior power of $K_1(\underline{f})$, and where $d_r:K_r(\underline{f}) \to K_{r-1}(\underline{f})$ is the unique sequence of $R$-module homomorphisms such that $d_1(g_1,\dots,g_n) = f_1g_1 + \dots + f_ng_n$ and such that $(K_{\bullet}(\underline{f}),d_\bullet)$ is a differential graded $R$-algebra. To be precise, $d_{r-1}\circ d_r$ equals the zero homomorphism, and $d_{r+s}(\alpha \wedge \beta) = d_r(\alpha)\wedge \beta + (-1)^r\alpha\wedge d_s(\beta)$ for every $r,s\geq 0$, for every $\alpha\in K_r(\underline{f})$, and for every $\beta\in K_s(\underline{f})$.

For every integer $r\geq 1$, denote by $Z_r(\underline{f})$ the kernel of $d_r$. Since $\underline{f}$ is a regular sequence, this is the same as the image of $d_{r+1}$. Thus, also define $Z_0(\underline{f})$ to be the image of $d_1$, i.e., the ideal $I$ generated by $\underline{f}$. In particular, $Z_{n+1}(\underline{f})$ is the zero module, and $Z_n(\underline{f})$ is $K_{n+1}(\underline{f})$.

**Fact 1.** For all $r$ with $1\leq r \leq n$, $Z_r(\underline{f})$ is a reflexive $S$-module.

**Proof.** By construction $K_r(\underline{f})$ is a free $S$-module, hence reflexive, and $K_{r-1}(\underline{f})$ is torsion-free (even free). The kernel of every $S$-module homomorphism from a reflexive module to a torsion-free module is reflexive. **QED**

**Fact 2.** For $M=Z_{n-1}(\underline{f})$, the first Fitting ideal $\text{Fitt}_1(M)$ equals the ideal $I$ generated by $(f_0,\dots,f_n)$.

**Proof.** The Fitting ideal can be computed from any finite presentation of $M$. For $Z_{n-1}(\underline{f})$, one such presentation is $d_{n+1}:K_{n+1}(\underline{f}) \to K_n(\underline{f})$. By self-duality of the Koszul complex, the Fitting ideal of $d_{n+1}$ is $I$. **QED**

Since the Fitting ideal is intrinsic, if $M$ is equivariant, then $I$ is a homogeneous ideal. However, there are many $S_+$-primary ideals $I$ that are generated by a regular sequence, yet are not homogeneous ideals. For instance, one example is $$\underline{f}=(x_0,x_1,\dots,x_{n-2},x_{n-1}-x_n^2,x_n^3).$$ For $n\geq 2$, for such $\underline{f}$, the module $M=Z_{n-1}(\underline{f})$ is reflexive, the restriction of $\widetilde{M}$ to $\mathbb{A}^{n+1}\setminus\{0\}$ is locally free (visibly it is locally free on each $D(f_i)$ for $i=0,\dots,n$). Yet $\widetilde{M}$ is not equivariant since $I$ is not a homogeneous ideal.

**Edit.** Song Sun asked the following variant of the question. For a reflexive $S$-module that is locally free on $\mathbb{A}^{n+1}\setminus\{0\}$ and whose Fitting ideals are all homogeneous, is the module equivariant? Also, note that every module as constructed above (also allowing other syzygy modules of the complexes) has rank $\geq n$. So here is a second variant: is every reflexive $S$-module that is locally free on $\mathbb{A}^{n+1}\setminus\{0\}$ equivariant provided that the rank is $\leq n-1$?

1more comment