To determine all values of $n$ for which the function $f$ is bijective, we need to analyze the behavior of the function $f$ defined on the set of all possible configurations of $n$ lamps. We will use polynomial theory and properties of finite fields to achieve this.

1. Polynomial Representation:
Consider the configuration of lamps as a polynomial $g_a(x)$ in $\mathbb{F}_2[x]$ modulo $x^n - 1$. Each lamp configuration can be represented as a polynomial where the coefficient of $x^i$ is 1 if the $i$-th lamp is on and 0 if it is off.

2. Function Definition:
The function $f$ can be expressed in terms of polynomials as:
$$f(g_a) = \frac{x g_a(x^2) - g_a(x)}{x-1} \pmod{x^n - 1}$$
This function describes the transformation of the lamp configuration after performing the cool procedure.

3. Bijectivity Condition:
For $f$ to be bijective, it must be injective. This means that if $f(g_a) = 0$, then $g_a$ must be 0. We need to analyze when $f(g_a) = 0$ implies $g_a = 0$.

4. **Factorization of $n$**:
Let $n = 2^k \cdot m$, where $m$ is odd. We will consider two cases: $m = 1$ and $m > 1$.

5. **Case $m > 1$:
- Lemma**: If $x^n - 1 \mid x g_a(x^2) - g_a(x)$, then:
$$\frac{x g_a(x^2) - g_a(x)}{x-1} = 0 \quad \text{or} \quad \frac{x^n - 1}{x-1} \pmod{x^n - 1}$$
- Proof of Lemma: Since $\frac{x^n - 1}{x-1}$ divides $\frac{x g_a(x^2) - g_a(x)}{x-1}$, we have:
$$\frac{x g_a(x^2) - g_a(x)}{x-1} = q(x) \cdot \frac{x^n - 1}{x-1}$$
for some polynomial $q(x)$. This implies that the lemma holds.

- Permutation Analysis: Consider the function $h(x) = 2x + 1 \pmod{n}$. This function is a permutation on the set $\{0, 1, \ldots, n-1\}$ with at least two cycles when $m > 1$. This is because $h(m \cdot 2^k - 1) = m \cdot 2^k - 1$, forming a cycle of length 1.

- Cycle Construction: For a cycle of length $j$ containing elements $c_1, c_2, \ldots, c_j$, construct the polynomial:
$$g_a(x) = \sum_{i=1}^j x^{c_i}$$
This polynomial satisfies $x^n - 1 \mid x g_a(x^2) - g_a(x)$. Since there are at least two cycles, either one of them is 0, or the sum of them will be 0. Thus, there exists a non-zero polynomial $g$ such that $f(g) = 0$.

6. **Case $m = 1$**:
- When $n = 2^k$, we need to show that $f$ is bijective. Consider the function $h(x) = 2x + 1 \pmod{n}$. If $g_a(x)$ has a term $x^m$ where $m \neq 2^k - 1$, then:
$$x^n - 1 \mid x g_a(x^2) - g_a(x)$$
By analyzing the $2$-adic valuation $\upsilon_2$, we can show that no such $m_1$ exists, leading to a contradiction. Therefore, $g_a(x) = x^{2^k - 1}$ gives $f(g_a) = \frac{x^{2^k} - 1}{x-1} \neq 0$.

Conclusion:
- For $n = 2^k$, the function $f$ is bijective.
- For $n = 2^k \cdot m$ with $m > 1$, the function $f$ is not bijective.

The final answer is $\boxed{ n = 2^k }$.