Here is a proof of the claim using results from Arthur's monograph The Endoscopic Classification of Representations: Orthogonal and Symplectic Groups.

Let $N = 2n$ be an even integer, and $\pi$ a cuspidal automorphic representation (henceforth: CAR) of $GL_{2n} / F$ satisfying $\pi^\vee \cong \pi$.

• Since $\pi \cong \pi^\vee$, the Rankin-Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. We have the factorisation $L(\pi \times \pi, s) = L(\pi, S^2, s) L(\pi, \wedge^2, s)$ and both factors are non-vanishing at $s = 1$, so exactly one of them must have a pole; thus $\pi$ is orthogonal or symplectic, in the sense of the question, but never both.
• The discussion preceding Theorem 1.5.3 of Arthur shows that $\pi$ defines an element of his set $\tilde{\Phi}_{\mathrm{sim}}(N)$ of global parameters, and this lies in the subset $\tilde{\Phi}_{\mathrm{sim}}(G)$ for a uniquely determined quasi-split group $G$ whose Langlands dual $\widehat{G}$ is either $SO_{2n}$ or $Sp_{2n}$.
• Theorem 1.5.3 of Arthur shows that $\widehat{G}$ is symplectic if $\pi$ is of symplectic type, and $\widehat{G}$ is orthogonal if $\pi$ is of orthogonal type. (This is a very deep theorem, despite sounding like a tautology!)
• Theorem 1.4.2 of op.cit. now shows that there is an automorphic representation $\sigma$ of $G$ such that, for every $v$, the Weil–Deligne representation associated to $\pi_v$ is the image in $GL_N$ of the ${}^L G$-valued parameter associated to $\sigma_v$. So, in particular, it is symplectic (resp. orthogonal) if $\pi$ is.

Here is a proof of the claim using results from Arthur's monograph The Endoscopic Classification of Representations: Orthogonal and Symplectic Groups.

Let $N = 2n$ be an even integer, and $\pi$ a cuspidal automorphic representation satisfying $\pi^\vee \cong \pi$.

• Since $\pi \cong \pi^\vee$, the Rankin-Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. We have the factorisation $L(\pi \times \pi, s) = L(\pi, S^2, s) L(\pi, \wedge^2, s)$ and both factors are non-vanishing at $s = 1$, so exactly one of them must have a pole; thus $\pi$ is orthogonal or symplectic, in the sense of the question, but never both.
• The discussion preceding Theorem 1.5.3 of Arthur shows that $\pi$ defines an element of his set $\tilde{\Phi}_{\mathrm{sim}}(N)$ of global parameters, and this lies in the subset $\tilde{\Phi}_{\mathrm{sim}}(G)$ for a uniquely determined quasi-split group $G$ whose Langlands dual $\widehat{G}$ is either $SO_{2n}$ or $Sp_{2n}$.
• Theorem 1.5.3 of Arthur shows that $\widehat{G}$ is symplectic if $\pi$ is of symplectic type, and $\widehat{G}$ is orthogonal if $\pi$ is of orthogonal type. (This is a very deep theorem, despite sounding like a tautology!)
• Theorem 1.4.2 of op.cit. now shows that there is a cuspidal automorphic representation $\sigma$ of $G$ such that, for every $v$, the Weil–Deligne representation associated to $\pi_v$ is the image in $GL_N$ of the ${}^L G$-valued parameter associated to $\sigma_v$. So, in particular, it is symplectic (resp. orthogonal) if $\pi$ is.

Here is a proof of the claim using results from Arthur's monograph The Endoscopic Classification of Representations: Orthogonal and Symplectic Groups.

Let $N = 2n$ be an even integer, and $\pi$ a cuspidal automorphic representation (henceforth: CAR) of $GL_{2n} / F$ satisfying $\pi^\vee \cong \pi$.

• Since $\pi \cong \pi^\vee$, the Rankin-Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. We have the factorisation $L(\pi \times \pi, s) = L(\pi, S^2, s) L(\pi, \wedge^2, s)$ and both factors are non-vanishing at $s = 1$, so exactly one of them must have a pole; thus $\pi$ is orthogonal or symplectic, in the sense of the question, but never both.
• The discussion preceding Theorem 1.5.3 of Arthur shows that $\pi$ defines an element of his set $\tilde{\Phi}_{\mathrm{sim}}(N)$ of global parameters, and this lies in the subset $\tilde{\Phi}_{\mathrm{sim}}(G)$ for a uniquely determined quasi-split group $G$ whose Langlands dual $\widehat{G}$ is either $SO_{2n}$ or $Sp_{2n}$.
• Theorem 1.5.3 of Arthur shows that $\widehat{G}$ is symplectic if $G$ is of symplectic type, and orthogonal if $\widehat{G}$ is of orthogonal type. (This is a very deep theorem, despite sounding like a tautology!)
• Theorem 1.4.2 of op.cit. now shows that there is an automorphic representation $\sigma$ of $G$ such that, for every $v$, the Weil–Deligne representation associated to $\pi_v$ is the image in $GL_N$ of the ${}^L G$-valued parameter associated to $\sigma_v$. So, in particular, it is symplectic (resp. orthogonal) if $\pi$ is.

Here is a proof of the claim using results from Arthur's monograph The Endoscopic Classification of Representations: Orthogonal and Symplectic Groups.

Let $N = 2n$ be an even integer, and $\pi$ a cuspidal automorphic representation (henceforth: CAR) of $GL_{2n} / F$ satisfying $\pi^\vee \cong \pi$.

• Since $\pi \cong \pi^\vee$, the Rankin-Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. We have the factorisation $L(\pi \times \pi, s) = L(\pi, S^2, s) L(\pi, \wedge^2, s)$ and both factors are non-vanishing at $s = 1$, so exactly one of them must have a pole; thus $\pi$ is orthogonal or symplectic, in the sense of the question, but never both.
• The discussion preceding Theorem 1.5.3 of Arthur shows that $\pi$ defines an element of his set $\tilde{\Phi}_{\mathrm{sim}}(N)$ of global parameters, and this lies in the subset $\tilde{\Phi}_{\mathrm{sim}}(G)$ for a uniquely determined quasi-split group $G$ whose Langlands dual $\widehat{G}$ is either $SO_{2n}$ or $Sp_{2n}$.
• Theorem 1.5.3 of Arthur shows that $\widehat{G}$ is symplectic if $\pi$ is of symplectic type, and $\widehat{G}$ is orthogonal if $\pi$ is of orthogonal type. (This is a very deep theorem, despite sounding like a tautology!)
• Theorem 1.4.2 of op.cit. now shows that there is an automorphic representation $\sigma$ of $G$ such that, for every $v$, the Weil–Deligne representation associated to $\pi_v$ is the image in $GL_N$ of the ${}^L G$-valued parameter associated to $\sigma_v$. So, in particular, it is symplectic (resp. orthogonal) if $\pi$ is.