According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ where $i$ is the inclusion $O(1)^2\subset O(2)$ and $tr$ is the transfer. The result is that it should be the sum with possibly signs of the identity and the switch map.

Now, if we compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2\rightarrow BO(1)\rightarrow Be$ where all unnamed arrows are transfers, we should get trivial map as the transfer associated to the inclusion $e\subset O(2)$ is trivial. This can be seen, for example, using Theorem II.17 of Feshbach "The transfer and compact Lie groups" Transactions of the American Mathematical Society vol 251 (1979) pp.139-169. So the two signs have to differ.

On the other hand, $Bi$ composed with the switch map is still $Bi$ up to homotopy since the conjugation by the permutation matrix $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ on $O(2)$ swithches the two factors of $O(1)$. Thus by considering the composition $BO(1)^2\stackrel{Switch}{\rightarrow} BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ we see that the two signs have to be same.

[Added afther the comment by Oscar Randal-Williams]

The map $2(id-Switch)$ should be non-trivial for the following reason.

Let's compute the induced maps in homology of infinite loop space by $2id$ and $2Switch$ The multiplication by $2$ induces in homology $F\circ V$ where $F$ is Frobenius, $V$ is Verschibung. Thus if we denote $e_i\otimes e_j$ the generator of $H_i(BO(1))\otimes H_j(BO(1))$, and we identify it with its image in $H_{i+j}(QBO(1)^2)$, then $\Omega ^{\infty}(2id) $ sends, let's say, $e_4\otimes e_2$ to $e_2^2\otimes e_1^2$, whereas $\Omega ^{\infty}(2Switch)$ sends $e_4\otimes e_2$ to $e_1^2 \otimes e_2^2$.

As $H_*(Q(BO(1)^2))$ is a polynomial algebra, those elements do differ.

[Added Nov.30] Here is another argument to show that $2id$ and $2Switch$ are different. As the Morava K-theory of $BO(1)^2$ is concentrated in even degrees, the Morava $E$-cohomology of $BO(1)^2$ is 2-torsion free, at least after $I_n$-adic completion (probably a simple 2-adic completion should suffice). Now, the identity and the switch map induce different maps on $E^*(BO(1)^2)$, so $2id$ and $2Switch$ induce different maps on $E^*(BO(1)^2)$ as well.

[Added Dec. 07 after a private communication with Oscar Randall-Williams]

What goes wrong in the above was the assumption that the "coefficients" are in $\{\pm 1\}$ whereas they lie in certain ring of units of an appropriate Burnside ring c.f. the comment by Oscar Randall-Williams below.

According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ where $i$ is the inclusion $O(1)^2\subset O(2)$ and $tr$ is the transfer. The result is that it should be the sum with possibly signs of the identity and the switch map.

Now, if we compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2\rightarrow BO(1)\rightarrow Be$ where all unnamed arrows are transfers, we should get trivial map as the transfer associated to the inclusion $e\subset O(2)$ is trivial. This can be seen, for example, using Theorem II.17 of Feshbach "The transfer and compact Lie groups" Transactions of the American Mathematical Society vol 251 (1979) pp.139-169. So the two signs have to differ.

On the other hand, $Bi$ composed with the switch map is still $Bi$ up to homotopy since the conjugation by the permutation matrix $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ on $O(2)$ swithches the two factors of $O(1)$. Thus by considering the composition $BO(1)^2\stackrel{Switch}{\rightarrow} BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ we see that the two signs have to be same.

[Added afther the comment by Oscar Randal-Williams]

The map $2(id-Switch)$ should be non-trivial for the following reason.

Let's compute the induced maps in homology of infinite loop space by $2id$ and $2Switch$ The multiplication by $2$ induces in homology $F\circ V$ where $F$ is Frobenius, $V$ is Verschibung. Thus if we denote $e_i\otimes e_j$ the generator of $H_i(BO(1))\otimes H_j(BO(1))$, and we identify it with its image in $H_{i+j}(QBO(1)^2)$, then $\Omega ^{\infty}(2id) $ sends, let's say, $e_4\otimes e_2$ to $e_2^2\otimes e_1^2$, whereas $\Omega ^{\infty}(2Switch)$ sends $e_4\otimes e_2$ to $e_1^2 \otimes e_2^2$.

As $H_*(Q(BO(1)^2))$ is a polynomial algebra, those elements do differ.

[Added Nov.30] Here is another argument to show that $2id$ and $2Switch$ are different. As the Morava K-theory of $BO(1)^2$ is concentrated in even degrees, the Morava $E$-cohomology of $BO(1)^2$ is 2-torsion free, at least after $I_n$-adic completion (probably a simple 2-adic completion should suffice). Now, the identity and the switch map induce different maps on $E^*(BO(1)^2)$, so $2id$ and $2Switch$ induce different maps on $E^*(BO(1)^2)$ as well.

[Added Dec. 07 after a private communication with Oscar Randall-Williams] (edited Dec 09)

What goes wrong in the above was the assumption that the "coefficients" are in $\{\pm 1\}$ whereas they lie in certain ring of units of an appropriate Burnside ring c.f. Mitchell-Priddy "A double coset formula for levi subgroups and splitting $BGL _n$" This issue is more ore less adressed in Remark (1.6), as well as in the remark following Theorem (2.3).

According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ where $i$ is the inclusion $O(1)^2\subset O(2)$ and $tr$ is the transfer. The result is that it should be the sum with possibly signs of the identity and the switch map.

Now, if we compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2\rightarrow BO(1)\rightarrow Be$ where all unnamed arrows are transfers, we should get trivial map as the transfer associated to the inclusion $e\subset O(2)$ is trivial. This can be seen, for example, using Theorem II.17 of Feshbach "The transfer and compact Lie groups" Transactions of the American Mathematical Society vol 251 (1979) pp.139-169. So the two signs have to differ.

On the other hand, $Bi$ composed with the switch map is still $Bi$ up to homotopy since the conjugation by the permutation matrix $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ on $O(2)$ swithches the two factors of $O(1)$. Thus by considering the composition $BO(1)^2\stackrel{Switch}{\rightarrow} BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ we see that the two signs have to be same.

[Added afther the comment by Oscar Randal-Williams]

The map $2(id-Switch)$ should be non-trivial for the following reason.

Let's compute the induced maps in homology of infinite loop space by $2id$ and $2Switch$ The multiplication by $2$ induces in homology $F\circ V$ where $F$ is Frobenius, $V$ is Verschibung. Thus if we denote $e_i\otimes e_j$ the generator of $H_i(BO(1))\otimes H_j(BO(1))$, and we identify it with its image in $H_{i+j}(QBO(1)^2)$, then $\Omega ^{\infty}(2id) $ sends, let's say, $e_4\otimes e_2$ to $e_2^2\otimes e_1^2$, whereas $\Omega ^{\infty}(2Switch)$ sends $e_4\otimes e_2$ to $e_1^2 \otimes e_2^2$.

As $H_*(Q(BO(1)^2))$ is a polynomial algebra, those elements do differ.

[Added Nov.30] Here is another argument to show that $2id$ and $2Switch$ are different. As the Morava K-theory of $BO(1)^2$ is concentrated in even degrees, the Morava $E$-cohomology of $BO(1)^2$ is 2-torsion free, at least after $I_n$-adic completion (probably a simple 2-adic completion should suffice). Now, the identity and the switch map induce different maps on $E^*(BO(1)^2)$, so $2id$ and $2Switch$ induce different maps on $E^*(BO(1)^2)$ as well.

According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ where $i$ is the inclusion $O(1)^2\subset O(2)$ and $tr$ is the transfer. The result is that it should be the sum with possibly signs of the identity and the switch map.

Now, if we compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2\rightarrow BO(1)\rightarrow Be$ where all unnamed arrows are transfers, we should get trivial map as the transfer associated to the inclusion $e\subset O(2)$ is trivial. This can be seen, for example, using Theorem II.17 of Feshbach "The transfer and compact Lie groups" Transactions of the American Mathematical Society vol 251 (1979) pp.139-169. So the two signs have to differ.

On the other hand, $Bi$ composed with the switch map is still $Bi$ up to homotopy since the conjugation by the permutation matrix $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ on $O(2)$ swithches the two factors of $O(1)$. Thus by considering the composition $BO(1)^2\stackrel{Switch}{\rightarrow} BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ we see that the two signs have to be same.

[Added afther the comment by Oscar Randal-Williams]

The map $2(id-Switch)$ should be non-trivial for the following reason.

Let's compute the induced maps in homology of infinite loop space by $2id$ and $2Switch$ The multiplication by $2$ induces in homology $F\circ V$ where $F$ is Frobenius, $V$ is Verschibung. Thus if we denote $e_i\otimes e_j$ the generator of $H_i(BO(1))\otimes H_j(BO(1))$, and we identify it with its image in $H_{i+j}(QBO(1)^2)$, then $\Omega ^{\infty}(2id) $ sends, let's say, $e_4\otimes e_2$ to $e_2^2\otimes e_1^2$, whereas $\Omega ^{\infty}(2Switch)$ sends $e_4\otimes e_2$ to $e_1^2 \otimes e_2^2$.

As $H_*(Q(BO(1)^2))$ is a polynomial algebra, those elements do differ.

[Added Nov.30] Here is another argument to show that $2id$ and $2Switch$ are different. As the Morava K-theory of $BO(1)^2$ is concentrated in even degrees, the Morava $E$-cohomology of $BO(1)^2$ is 2-torsion free, at least after $I_n$-adic completion (probably a simple 2-adic completion should suffice). Now, the identity and the switch map induce different maps on $E^*(BO(1)^2)$, so $2id$ and $2Switch$ induce different maps on $E^*(BO(1)^2)$ as well.

[Added Dec. 07 after a private communication with Oscar Randall-Williams]

What goes wrong in the above was the assumption that the "coefficients" are in $\{\pm 1\}$ whereas they lie in certain ring of units of an appropriate Burnside ring c.f. the comment by Oscar Randall-Williams below.

According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ where $i$ is the inclusion $O(1)^2\subset O(2)$ and $tr$ is the transfer. The result is that it should be the sum with possibly signs of the identity and the switch map.

Now, if we compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2\rightarrow BO(1)\rightarrow Be$ where all unnamed arrows are transfers, we should get trivial map as the transfer associated to the inclusion $e\subset O(2)$ is trivial. This can be seen, for example, using Theorem II.17 of Feshbach "The transfer and compact Lie groups" Transactions of the American Mathematical Society vol 251 (1979) pp.139-169. So the two signs have to differ.

On the other hand, $Bi$ composed with the switch map is still $Bi$ up to homotopy since the conjugation by the permutation matrix $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ on $O(2)$ swithches the two factors of $O(1)$. Thus by considering the composition $BO(1)^2\stackrel{Switch}{\rightarrow} BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ we see that the two signs have to be same.

[Added afther the comment by Oscar Randal-Williams]

The map $2(id-Switch)$ should be non-trivial for the following reason.

Let's compute the induced maps in homology of infinite loop space by $2id$ and $2Switch$ The multiplication by $2$ induces in homology $F\circ V$ where $F$ is Frobenius, $V$ is Verschibung. Thus if we denote $e_i\otimes e_j$ the generator of $H_i(BO(1))\otimes H_j(BO(1))$, and we identify it with its image in $H_{i+j}(QBO(1)^2)$, then $\Omega ^{\infty}(2id) $ sends, let's say, $e_4\otimes e_2$ to $e_2^2\otimes e_1^2$, whereas $\Omega ^{\infty}(2Switch)$ sends $e_4\otimes e_2$ to $e_1^2 \otimes e_2^2$.

As $H_*(Q(BO(1)^2))$ is a polynomial algebra, those elements do differ.

According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ where $i$ is the inclusion $O(1)^2\subset O(2)$ and $tr$ is the transfer. The result is that it should be the sum with possibly signs of the identity and the switch map.

Now, if we compute the composition $BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2\rightarrow BO(1)\rightarrow Be$ where all unnamed arrows are transfers, we should get trivial map as the transfer associated to the inclusion $e\subset O(2)$ is trivial. This can be seen, for example, using Theorem II.17 of Feshbach "The transfer and compact Lie groups" Transactions of the American Mathematical Society vol 251 (1979) pp.139-169. So the two signs have to differ.

On the other hand, $Bi$ composed with the switch map is still $Bi$ up to homotopy since the conjugation by the permutation matrix $\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)$ on $O(2)$ swithches the two factors of $O(1)$. Thus by considering the composition $BO(1)^2\stackrel{Switch}{\rightarrow} BO(1)^2\stackrel{Bi}{\rightarrow} BO(2)\stackrel{tr}{\rightarrow} BO(1)^2$ we see that the two signs have to be same.

[Added afther the comment by Oscar Randal-Williams]

The map $2(id-Switch)$ should be non-trivial for the following reason.

Let's compute the induced maps in homology of infinite loop space by $2id$ and $2Switch$ The multiplication by $2$ induces in homology $F\circ V$ where $F$ is Frobenius, $V$ is Verschibung. Thus if we denote $e_i\otimes e_j$ the generator of $H_i(BO(1))\otimes H_j(BO(1))$, and we identify it with its image in $H_{i+j}(QBO(1)^2)$, then $\Omega ^{\infty}(2id) $ sends, let's say, $e_4\otimes e_2$ to $e_2^2\otimes e_1^2$, whereas $\Omega ^{\infty}(2Switch)$ sends $e_4\otimes e_2$ to $e_1^2 \otimes e_2^2$.

As $H_*(Q(BO(1)^2))$ is a polynomial algebra, those elements do differ.

[Added Nov.30] Here is another argument to show that $2id$ and $2Switch$ are different. As the Morava K-theory of $BO(1)^2$ is concentrated in even degrees, the Morava $E$-cohomology of $BO(1)^2$ is 2-torsion free, at least after $I_n$-adic completion (probably a simple 2-adic completion should suffice). Now, the identity and the switch map induce different maps on $E^*(BO(1)^2)$, so $2id$ and $2Switch$ induce different maps on $E^*(BO(1)^2)$ as well.