I'm having a hard time trying to prove this, but somehow it seems true... The problem is:
Let $\mathbb{S}^m$ be the space of all real symmetric matrices, and let $U\in \mathbb{R}^{m\times m}$ be an arbitrary orthogonal matrix. Also, let $L\colon \mathbb{S}^m\to \mathbb{R}^n$ be a linear operator, and let $X$ be a subspace of $\mathbb{S}^m$.
Is $\dim(L(U^\top XU))=\dim(L(X))$?
This is the general form of my question, but the operator I'm interested in has the form $$M\mapsto L(M)\doteq \begin{bmatrix} \langle L_1, M\rangle, \ldots, \langle L_n, M\rangle \end{bmatrix}$$ where $L_1,\ldots,L_n\in \mathbb{S}^m$, which is just an adjoint of another linear operator written in the canonical basis of $\mathbb{R}^n$. However, $L$ is not assumed to be injective nor surjective.
Note that every operator from $\Bbb S^m$ to $\Bbb R^n$ can be written in the form of your particular map $L(M)$ for a certain choice of $L_1,\dots,L_n \in \Bbb S^m$. So, there is no difference between the "general" form of your question and the consideration of maps of the form you describe.
With that said, the answer to the question is no. As an example, define $L:\Bbb S^2 \to \Bbb R$ by $$ L(M) = m_{11}. $$ Let $X$ denote the one-dimensional subspace spanned by the matrix $$ X_0 = \pmatrix{1&0\\0&-1}, $$ and let $U$ denote the orthogonal matrix $$ U = \frac 1{\sqrt{2}}\pmatrix{1&1\\1&-1}. $$ Note that $$ Y_0 := U^TX_0U = \pmatrix{0&1\\1&0}. $$ We see that $L(U^TXU)$ is the span of $L(Y_0) = 0$. So, $$ \dim L(X) = 1 \neq \dim(L(U^TXU)) = 0. $$