Do automorphisms of a finitely generated free module restrict to automorphisms on every invariant submodule?

If $R$ is Noetherian, this is true. Notice that $\gamma(N)\subset N$ implies $N\subset \gamma^{-1} N$. Thus, we get a chain $N\subset\gamma^{-1} N\subset \gamma^{-2} N\subset\cdots$ and by Noetherian hypothesis, we have $\gamma^{-n}(N)=\gamma^{-n-1}(N)$ for some $n$ and then it is easy to see that $N=\gamma(N)$, proving what you need.

The answer is yes in all cases, even if $N$ is not finitely generated.

The case where $N$ is finitely generated is settled by Orzech's theorem; see this question. (To reduce to said theorem, pass to $\gamma^{-1}$ instead of $\gamma$, which restricts to a surjective $R$-module homomorphism $N \supseteq \gamma(N) \to N$.) This question was answered by user26857, the very same who already observed that the result holds if $N$ is only assumed to be finitely generated.

The general case is settled by using the Cayley–Hamilton theorem. I follow the approach taken by Darij Grinberg in this writing, which was also linked in his own answer to the aforementioned question regarding Orzech's theorem. The key ingredient to Darij's writing is the following lemma (presented here in my own words):

Lemma 0.5. ($A$ is a commutative ring with unit and $n\in\mathbb{N}_0$ is given.) Let $g : A^n \to A^n$ be an $A$-linear map and let $V \subseteq A^n$ be a submodule satisfying $g^{-1}(V) \subseteq V$. Then we have $g(V) \subseteq V$.

Proof. The case $n = 0$ is trivial, but needs to be considered separately. For the remainder of this proof, we assume $n \geq 1$. By the Cayley–Hamilton theorem, we may choose some $c_0,\ldots,c_n\in A$ with $c_n = 1$ such that $$ c_ng^n + c_{n-1}g^{n-1} + \cdots + c_1g + c_0 = 0.\tag*{(1)} $$ We prove by induction that every $u\in \{0,1,\ldots,n\}$ satisfies $$ \big(c_ng^{n - u} + c_{n-1}g^{n - u - 1} + \cdots + c_{u+1}g + c_u\big)(V) \subseteq V.\tag*{(2)} $$

• The base case $u = 0$ is clear from $(1)$.
• Let $p \in \{1,\ldots,n\}$ be given and suppose that $(2)$ holds for $u = p - 1$. We must now prove that it also holds for $u = p$. To that end, let $v\in V$ be given. By the induction hypothesis, we have $$ \big(c_ng^{n - p + 1} + c_{n-1}g^{n - p} + \cdots + c_pg + c_{p-1}\big)(v) \in V. $$ Since we also have $c_{p-1}v \in V$, it follows that \begin{align*} &\: g\Big(\big(c_ng^{n - p} + c_{n-1}g^{n - p - 1} + \cdots + c_{p+1}g + c_p\big)(v)\Big)\\[1ex] =& \: \Big(g\big(c_ng^{n - p} + c_{n-1}g^{n - p - 1} + \cdots + c_{p+1}g + c_p\big)\Big)(v)\\[1ex] =& \: \big(c_ng^{n - p + 1} + c_{n-1}g^{n - p} + \cdots + c_{p+1}g^2 + c_pg\big)(v) \quad \in V. \end{align*} Therefore we have $$ \big(c_ng^{n - p} + c_{n-1}g^{n - p - 1} + \cdots + c_{p+1}g + c_p\big)(v) \in g^{-1}(V) \subseteq V, $$ which shows that $(2)$ holds for $u = p$ as well.

The proof shows that $(2)$ holds for all $u \in \{0,1,\ldots,n\}$, so in particular it holds for $u = n - 1$ (here we use the assumption $n \geq 1$). In other words, we have $$ \big(c_ng + c_{n-1}\big)(V) \subseteq V. $$ Recall that $c_n = 1$ holds; therefore we find $g(V) \subseteq V$.$\quad\blacksquare$

Observe that the lemma cannot be reversed: there are ample examples where $g(V) \subseteq V$ holds but $g^{-1}(V) \not\subseteq V$. (Let $V = \langle x\rangle \subsetneq A$ be a proper principal ideal and let $g : A \to A$ be multiplication with $x$.)

To solve the problem at hand, we apply the above lemma to $g = \gamma^{-1}$ and $V = N$.