The answer is **no**: a linear transformation of $F$ which commutes with $\phi$ is an automorphism of $F$.

This is a seemingly inelegant but simple argument.

Any map $\psi: F \rightarrow F$ can be represented uniquely as a polynomial $P_\psi (x) = \sum a_i x^i$ of degree less than $2^n$, with each $a_i \in F$. If the map $\psi$ is linear, then in fact $a_i =0$ for all $i\neq 2^j$, so that the degree of $P_\psi$ is in fact at most $2^{n-1}$. Indeed, there are $2^{n^2}$ linear transformations of $F$ and the same number of polynomials of the form $\sum_{0 \le j \le
n - 1}b_jx^{2^j}$, all of which induce linear transformations.

Let $Q_\psi(x) = x^{2^{n-1}} P_\psi(\frac{1}{x})$ as polynomials; in other words, $Q_\psi$ is the "reverse" of $P_{\psi}$. Note that the degree of $Q_{\psi}$ is also at most $2^{n-1}-1$, because the coefficient of $1$ in $P_\psi$ is $0$.

Your condition is equivalent to the equation $(P_\psi(x))^{-1} = P_\psi(x^{-1})$ in value for any $x \neq 0$. Multiplying it out so everything is a polynomial, it's equivalent to $P_\psi(x)Q_\psi(x) = x^{2^{n-1}}$ in value for all $x$ including $0$. But two polynomials are equal in value iff they are equivalent modulo $x^{2^n}-x$. Both of these polynomials are of smaller degree than $2^n$, so they must in fact be equal. Unique factorization therefore says that $P_{\psi}$ must be a monomial, with coefficient $a$ such that $a^2=1$. As we are in characteristic $2$, this is unique: $a=1$.

I wouldn't be surprised if there were a more elegant answer based on automorphisms, but I don't see it immediately.

I figured out an answer based on multiplication-preserving endomorphisms that works generally in any characteristic *other* than $2$, and for any finite field (or any field in which every element is a square) in characteristic $2$.

The commutation equation says that for any $x$, we have $\psi(x^{-1}) = \psi(x)^{-1}$. Assume $x, y, x + y \neq 0$. Let's apply it to $x^{-1} + y^{-1}$:

$\psi((x^{-1} + y^{-1})^{-1}) = (\psi(x^{-1} + y^{-1}))^{-1}$

$\psi(\frac{xy}{x + y}) = (\psi(x^{-1}) + \psi(y^{-1}))^{-1}$ (using linearity)

$\psi(\frac{xy}{x + y}) = (\psi(x)^{-1} + \psi(y)^{-1})^{-1} = \frac{\psi(x) \psi(y)}{\psi(x) + \psi(y)}$

Multiplying it out, we get:

$\psi(\frac{xy}{x + y}) (\psi(x + y)) = \psi(x) \psi(y)$

Choose $y = y' - x$. Our conditions become $x, y', y'-x \neq 0$. Then

$\psi(\frac{x (y' - x)}{y'}) \psi(y') = \psi(x) \psi(y' - x)$

Using linearity and canceling terms, we get:

$\psi(\frac{x^2}{y'})\psi(y') = \psi(x)^2$

From here, this separates into two proofs: one for characteristic $2$ fields, and one for any field not of characteristic $2$.

For characteristic $2$ fields:

$\psi(z z')^2 \psi(1)^2 = \psi(z^2) \psi(1) \psi(z'^2) \psi(1) = \psi(z)^2 \psi(z')^2$ for any $z, z'$.

In fields of characteristic $2$, square roots are unique, if they exist. Therefore:

$\psi(z z') \psi(1) = \psi(z) \psi(z')$

For characteristic not $2$:

$\psi(x^2)\psi(1) = \psi(x)^2$ for any $x \neq 0, 1$. However, for both of those cases, this is obvious, so this is true for any $x$.

Let $x = z + z'$. Then:
$\psi((z + z')^2) \psi(1) = \psi(z + z') \psi(z + z')$. We can again cancel terms and divide by $2$ (except in characteristic $2$) to get:

$\psi(z z') \psi(1) = \psi(z) \psi(z')$.

Here, the proofs reunite.

In other words, $\psi$ must be a multiplication-preserving morphism up to scaling by $\psi(1)$. But $\psi(1) = \psi(1^{-1}) = \psi(1)^{-1}$, so $\psi(1) = \pm 1$. In characteristic $2$, these are the same, and the only answers are multiplication-preserving morphisms; more generally, negation (and composing negation with a multiplication-preserving morphism) is also possible.