Is self-adjointness really a property of an operator, or of an operator and an inner product?

Self-adjointness is the operator-theoretic version of symmetry of bilinear forms. The notion of a symmetric bilinear form can be defined without reference to any extra structure but in order to talk about self-adjoint operators, we need some mechanism to convert operators to bilinear forms. An inner product $g$ allows us to do just that. Namely, let $V$ be a finite dimensional real vector space and let $g$ be an inner product. Given a linear map $T$, we can define the bilinear form associated to $T$ to be $\beta^g_T(v,w) := g(Tv,w)$. This gives us a map (in fact, a linear isomorphism) from $\operatorname{End}(V) \cong V^{*} \otimes V$ to $\operatorname{Bil}(V \times V, \mathbb{R}) \cong V \otimes V$.

The operator $T$ is $g$-self-adjoint if the associated bilinear form $\beta^g_T$ is symmetric. By changing $g$ we get different different maps between $\operatorname{End}(V)$ and $\operatorname{Bil}(V \times V, \mathbb{R})$ so $T$ might be $g$-self-adjoint but not $g'$-self-adjoint. This is best seen by invoking the spectral theorem.

In the real case, we know that $T$ is $g$-self-adjoint if and only if $T$ is $g$-orthogonally diagonalizable. Denote by $\lambda_1, \dots, \lambda_k$ the distinct eigenvalues of $T$. Then $V = \bigoplus_{i=1}^k \ker(T - \lambda_i I)$ is a direct sum orthogonal decomposition of $V$. If we change the inner product $g$ to an inner product $g'$ such that some $\ker(T - \lambda_i I)$ won't be $g'$-orthogonal to some $\ker(T - \lambda_j I)$ (with $i \neq j$) then by the spectral theorem, $T$ won't be self-adjoint. This is always possible as long as $k > 1$. In other words, the only operators $T$ which are self-adjoint with respect to all inner products are the scalar operators.

In fact, a real operator $T$ is $g$-self-adjoint with respect to some inner product $g$ if and only if $T$ is diagonalizable. If $T$ is self-adjoint, it is orthogonally diagonalizable and in particular diagonalizable. For the other direction, if $T$ is diagonalizable we can always take an arbitrary basis of eigenvectors of $T$ and define an inner product $g$ by declaring this basis to be orthonormal - then by the spectral theorem $T$ will be $g$-orthogonal.