The major point here is that, for an operator $S$ on a Banach space (or Hilbert space) $X$, the number $\sup_{x \in X \setminus\{0\}} \frac{\|Sx\|}{\|x\|}$ is not the spectral radius of $S$ but the operator norm. The operator norm is always $\ge$ the spectral radius, but we cannot expect equality in general.
On a Hilbert space, one sufficient condition for equality of the operator norm and the spectral radius is that the operator be self-adjoint or, more generally, normal.
But as pointed out in a comment by Mateusz Kwaśnicki, if $T$ is self-adjoint on $L^2$, this does not imply that $T$ is self-adjoint on $H^1_0$ since the inner product there is different from the inner product on $L^2$.
Here is a concrete counterexample:
Let $\Omega = (0,2\pi)$ (endowed with the non-normalised Lebesgue measure) and define $z,v \in H^1_0 := H^1_0(\Omega)$ by
\begin{align*}
z(x) & = \frac{|\sin(x)|}{\sqrt{\pi}}, \\
v(x) & = \sin(\frac{1}{2}x)
\end{align*}
for all $x \in (0,2\pi)$.
We define the operator $T$ on $L^2$ by
$$
Tf = \langle f, z\rangle_{L^2} \cdot z
$$
for all $f \in L^2$. Then $T$ is a self-adjoint rank-$1$ projection on $L^2$ whose norm and spectral radius are thus equal to $1$. Clearly, the range of $T$ is a subset of $H^1_0$.
The restriction of $T$ to $H^1_0$ is again a non-zero projection and thus still has spectral radius $1$. But the operator norm of $T$ on $H^1_0$ is strictly larger than $1$. Indeed, we have
$$
\|T\|_{H^1_0 \to H^1_0} \ge \frac{\|Tv\|_{H^1_0}}{\|v\|_{H^1_0}} = \frac{\sqrt{512}}{\sqrt{45}\pi} > 1
$$
(we need to compute a few integrals to obtain the equality in the middle, but the computations are rather straightforward).
This proves that the operator norms of $T$ on $L^2$ and on $H^1_0$ are distinct, although the spectral radius on both spaces is $1$. In particular, $T$ cannot be self-adjoint (and not even normal) on $H^1_0$.
EDIT: An additional observation. While, in the example above, equality of the spectral radii on both spaces follows from the fact that $T$ acts as a projection on both spaces, I thought it might be worthwhile to point out that the equality of the spectral radii is actual a general fact:
Proposition. (Equality of spectral radii) Let $V,X$ be complex Banach spaces such that $V$ is continuously embedded in $X$. Let $T: X \to X$ be a bounded linear operator such that $TX \subseteq V$. Then the spectral radius of the operator $T: X \to X$ coincides with the spectral radius of the restriction $T|_V: V \to V$.
Proof. We use the spectral radius formula
$$
(*) \qquad r(S) = \lim_{n \to \infty} \|S^n\|^{1/n}
$$
which holds for the spectral radius $r(S)$ of each bounded linear operator $S$ on a complex Banach space.
For each $n \in \mathbb{N}$ the operator $(T|_V)^n = (T^n)|_V: V \to V$ factorizes as
$$
V \hookrightarrow X \overset{T^{n-1}}{\longrightarrow} X \overset{T}{\longrightarrow} V,
$$
so the spectral radius formula $(*)$ implies that $r(T|_V) \le r(T)$. But conversely, the operator $T^n: X \to X$ factorizes as
$$
X \overset{T}{\longrightarrow} V \overset{(T|_V)^{n-1}}{\longrightarrow} V \hookrightarrow X,
$$
so the spectral radius formula $(*)$ also implies that $r(T) \le r(T|_V)$. This proves the proposition.
Remark. What is quite nice about the proposition above is that it does not rely on eigenvalues, so no compactness assumption on the operator $T$ is needed.
