The Hessian of invariant functions on a Lie group Assume  that $G$  is  a  Lie  group with Lie  algebra $\mathfrak{g}$. We  fix an invariant Riemannian metric on $G$ and fix its corresponding $LC$  connection.
Consider the  natural  right action of $G$ on its  Lie  algebra $\mathfrak{g} \simeq \{X \in \chi^{\infty}({G}) \mid R_{g}^{*} X=X\}$, the  space of smooth vector fields which are invariant under right multiplications.
In fact the right action is defined as follows:
For  $g\in G$  and  $X\in  \mathfrak{g}$ define $X.g=L_{g}^{*} X$  where $L_{g}$ is the left  multiplication by $g$.
So  there is  a  natural (component wise)  action of  $G$ on $\mathfrak{g} \times \mathfrak{g}$
A smooth  function  $f:G \to \mathbb{R}$  is  called $G$-invariant if $f(g^{-1}hg)=f(h)$  for  all $g,h \in G$. For  example $Det:  Gl(n, \mathbb{R}) \to \mathbb{R}$.
A $2$-  linear map $T: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ is  $G$-invariant if $T((v,w).g)=T(v,w)$
Example  The $2$-linear  map $tr(u)tr(v)-tr(uv)$ defined on $M_{n}(\mathbb{R}) \times M_{n}(\mathbb{R})$ is a $Gl(n, \mathbb{R})$  invariant  map, with the natural (conjugate)  action of  $Gl(n, \mathbb{R})$ on its  Lie algebra  $M_{n}(\mathbb{R})$ as  described above.
Recall that for  a  Riemannian manifold  with the  corresponding $LC$  connection  $\nabla$, the  Hessian of a function $f$ defined on the manifold, is a two linear  map on the tangent space  with the formula $Hess (f).(V, W)=\nabla ^ {\nabla f}_{V}.W$

Question: Let $G$ be  a  Lie  group and  $f:G \to \mathbb{R}$  be  a $G$-invariant smooth function. Is its  Hessian  $Hess(f)$ a $G$-invariant $2$-linear map on the Lie algebra $\mathfrak{g}$ of $G$?

This question is  motivated by the following post.
Is there an explicit formula for the hessian of "Determinant"?
 A: Yes.  
In what follows, I use standard notation for the derivative; see, e.g., $\S$2.3 of Banach Spaces and Differential Calculus (Chapter 2) of the book referenced below.
First Derivative of $f$.
As the OP stated, a function $f: G \to \mathbb{R}$ is $G$ invariant means that 
$$
f(A) = f(g A g^{-1}) \quad \forall A, g \in G \tag{$\star$}
$$  If we differentiate this relation once we obtain
$$
D f(A) \cdot B_1 = \frac{d}{d\lambda} \left. f( g (A+\lambda B_1) g^{-1} ) \right|_{\lambda=0} = D f( g A g^{-1} ) \cdot g B_1 g^{-1} \tag{$\star \star$}
$$ where $B_1 \in T_A G$.  In the special case of the determinant on $G = Gl(n,\mathbb{R})$, this is simply saying that: 
$$
D \det(A) \cdot B_1 = \det(A) \operatorname{trace}(A^{-1} B_1) = \det( g A g^{-1} ) \operatorname{trace}(g A^{-1} g^{-1} g B_1 g^{-1} ) 
$$ which follows from basic properties of the determinant and trace of a matrix.
Second Derivative of $f$.
Now differentiate ($\star \star$) to obtain: 
$$
D^2 f(A) \cdot (B_1, B_2) =  D^2 f( g A g^{-1} ) \cdot ( g B_1 g^{-1}, g B_2 g^{-1} ) \tag{$\star \star \star$}
$$ where $B_1, B_2 \in T_A G$.  In the special case of the determinant on $G = Gl(n,\mathbb{R})$,  \begin{align*}
D^2 \det(A) \cdot (B_1, B_2) &=  \frac{d}{d \lambda} \left. D \det(A+\lambda B_2) \cdot B_1  \right|_{\lambda=0} \\
&= \frac{d}{d \lambda} \left. \det (A + \lambda B_2) \operatorname{trace} ( (A+\lambda B_2)^{-1} B_1 ) \right|_{\lambda=0} \\
 &= \det(A) \operatorname{trace} (A^{-1} B_2) \operatorname{trace}(A^{-1} B_1) - \det(A) \operatorname{trace}(A^{-1} B_2 A^{-1} B_1 )
\end{align*} which is clearly $G$-invariant in the sense of relation ($\star \star \star$).
Higher Derivatives of $f$.
This result for the first and second derivative seems to be true for higher derivatives of a $G$-invariant function.  In other words, differentiation preserves $G$-invariance in the sense given above.
Reference
For background info see, e.g., An Introduction to Lie groups Chapter 5 of
Jerrold E. Marsden, Tudor Ratiu, and Ralph Abraham.  Manifolds, Tensors, and Applications. Second Edition. Vol. 75. Springer Science & Business Media, 2012.
