Edit: the discussion is relevant only when the domain and the range of the linear map are the same, otherwise we cannot talk about similarity, regards to Christoph for the remark.
In general case - no, they are not equal. Let us declare a linear transformation $f: R^2 \to R^2$ and $g: R^2 \to R^2$.
Let us fix 2 bases in $R^2: B_{1} = \{(1, -1)^T, (2, 3)^T\}$ and $B_{2} = \{(4, 6)^T, (3, 1)^T\}$ (they constitute a basis, because they are linearly independent and there are 2 of them, which is $dim(R^2)$).
Let us say that linear transformations $f$ and $g$ have the same matrix with respect to bases $B_{1}$ and $B_{2}$ respectively:
$[f]_{B_{1}} = [g]_{B_{2}}= \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
We would need the following property on which you can read more in the Change of basis topic (or here) while discussing linear maps. This is a corollary of a more general theorem for the linear maps which have the same domain and range:
Assume that $f : R^n \to R^n$ is a linear map and let $B=\{v_{1},...,v_{n}\}$ and $C=\{v'
_{1},...,v'_{n}\}$ be bases of $R^n$. Let $g : R^n \to R^n$ be the uniquely determined linear map for which $g(v_{j}) = v'_{j}$ holds for every $1 ≤ j ≤ n.$ Then $[f]_{C} = [g]^{−1}_{B}[f]_{B}[g]_{B}$. Likewise $[f]_{B} = [g]_{B}[f]_{C}[g]_{B}^{-1}$.
We will show that the linear maps defined above have different matrices with respect to the standard basis $\{(1,0)^T, (0, 1)^T\}$ in $R^2$.
- $[f]$
We need matrix of linear map $h$ which maps standard basis in $R^2$ to the basis $B_{1}$
$[h] = \begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}$
$[h]^{-1} = \frac{1}{5}*\begin{bmatrix}3 & -2\\1 & 1\end{bmatrix}$
$[f] = [h][f]_{B_{1}}[h]^{-1} = \frac{1}{5}*\begin{bmatrix}7 & -3\\8 & -7\end{bmatrix}$
- $[g]$
As in the first case, we need matrix of linear map $h$ which maps standard basis in $R^2$ to the basis $B_{2}$
$[h] = \begin{bmatrix}4 & 3\\6 & 1\end{bmatrix}$
$[h]^{-1} = \frac{1}{14}*\begin{bmatrix}-1 & 3\\6 & -4\end{bmatrix}$
$[g] = [h][g]_{B_{2}}[h]^{-1} = \frac{1}{2}*\begin{bmatrix}3 & -1\\5 & -3\end{bmatrix}$
As we can see, matrices of linear maps $f$ and $g$ w.r.t. the standard basis are not equal, that is, these maps are not the same.